Kinematics 2D Problem Solving
Techniques
2023-09-09
www.njctl.org
https://njctl.org/video/?v=E-65h0F3kH8
Table of Contents:
Kinematics Problem Solving Techniques
Click on the topic to go to that section.
· A Ball Rolling Of a Table
· A Projectile Launched at an Angle from the Ground
· Dropped Package
· Cannon Ball Flight
· Fire Fighter and a Fire
· A Motor Boat Crossing a River
A Ball Rolling Off a Table
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A Ball Rolling Off a Table
A ball rolls off a 1.5 m tall horizontal table and lands on the
floor 0.70 m away.
a. How much time is the ball in the air?
b. How does that time compare with the time it takes for a dropped
ball to fall that same distance?
c. What is the ball’s velocity while it was on the table top?
d. What is the horizontal component of its velocity just prior to
impact?
e. What is the vertical component of its velocity just prior to impact?
f. What is the magnitude of its velocity just prior to impact?
g. What is the direction of its velocity just prior to impact?
A Ball Rolling Off a Table
a. How much time is the ball in the air?
Use the second kinematics equation in the y-direction to derive an
equation for the time it took for the ball to reach the ground.
Givens:
A Ball Rolling Off a Table
b. How does that time compare with the time it takes for a dropped
ball to fall that same distance?
The first ball "rolls off horizontally" (meaning it only has a horizontal
component of velocity) while the second ball is dropped (meaning it
was released from rest), so that the vertical component of the
velocity of both balls is zero.
Because the balls start from the same height above the ground,
have the same initial vertical component of velocity, and since air
resistance is negligible, both balls undergo the same, constant
acceleration due to gravity (-9.8 m/s2), the two balls will take the
same amount of time to reach the ground.
The times are the same.
A Ball Rolling Off a Table
c. What is the ball’s velocity while it was on the table top?
Because air resistance is negligible, after the ball leaves the
tabletop, the acceleration in the horizontal direction is zero,
meaning the ball travels with a constant horizontal velocity.
Since the horizontal velocity of the ball remains constant, apply
the second kinematics equation in the x-direction and solve for
v0x the initial horizontal component of velocity (which is the
velocity of the ball on the tabletop).
A Ball Rolling Off a Table
c. What is the ball’s velocity while it was on the table top?
Givens:
A Ball Rolling Off a Table
d. What is the horizontal component of its velocity just prior to
impact?
Just prior to impact, the horizontal component of velocity will
remain constant at 1.3 m/s since there is no horizontal
acceleration of the ball.
A Ball Rolling Off a Table
e. What is the vertical component of its velocity just prior to
impact?
Use the first kinematics equation in the y-direction to calculate the
vertical component of velocity just prior to impact. Setup the
variables to start at the moment the ball leaves the table to the
moment it hits the ground. Since the ball is rolled off horizontally,
the initial vertical velocity is zero. The total time in the air was
found in part (a).
A Ball Rolling Off a Table
f. What is the magnitude of its velocity just prior to impact?
Prior to impact, there are two components of the net velocity
which must be considered: horizontal and vertical. Apply the
Pythagorean Theorem to perform the vector addition.
The horizontal component of velocity is a constant, so use the
value calculated in part (c) for vx and the value from part (e) for
the vertical component vy.
vx
v
vy
A Ball Rolling Off a Table
g. What is the direction of its velocity just prior to impact?
When solving for the velocity, a sketch can be helpful, in addition
to using these equations (to solve for both the magnitude and
direction of the net velocity vector):
vx
v
vy
A negative angle means that the ball impacts at an angle of
76 degrees below the horizontal.
Projectile Launched at
an Angle from the
Ground
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Projectile Launched at Angle
from the Ground
A projectile is fired with an initial speed of 40 m/s at an angle of
23 degrees above the horizontal.
a. Determine the total time in the air.
b. Determine the maximum height reached by the projectile.
c. Determine the maximum horizontal distance covered by the
projectile.
d. Determine the velocity of the projectile 2 s after firing.
e. Suppose a second projectile is launched but at an angle of 67
degrees above the horizontal with the same initial speed. Without
using equations, compare the following for the first and second
projectile launches:
i. maximum height
ii. speed at the apex
iii. landing speed
Projectile Launched at Angle
from the Ground
a. Determine the total time in the air.
Make a sketch.
Time in the air is governed by motion in the y-direction. Take the
vertical component of the velocity.
Projectile Launched at Angle
from the Ground
a. Determine the total time in the air (cont).
Use the second kinematics equation to derive an expression for the
total time the projectile is in the air. Since the projectile begins and
ends on the ground, y0 and y are zero.
Givens:
factor out t
giving the
solution, t = 0;
which is when
the projectile
is launched.
Projectile Launched at Angle
from the Ground
b. Determine the maximum height reached by the projectile.
One approach is to use the third kinematics equation in the ydirection with the starting position on the ground and ending
position at the maximum height. When the projectile reaches
the maximum height its vertical velocity will be 0 m/s.
Givens:
Projectile Launched at Angle
from the Ground
c. Determine the maximum horizontal distance covered by the
projectile.
Use the second kinematics equation in the x-direction to derive an
expression for the maximum horizontal distance covered by the
projectile. Use the total time in the air in part (a).
Givens:
v0x = v0cosθ
Projectile Launched at Angle
from the Ground
d. Determine the velocity of the projectile 2 s after firing.
Below is a sketch of the projectile's trajectory. Since it takes 3.2 s
to land, it takes 1.6 s to be at its peak value. The projectile must be
on its way back down at 2 s. This will help you determine if our
result is reasonable later.
x(m)
v
1
2
3
t(s)
Projectile Launched at Angle
from the Ground
d. Determine the velocity of the projectile 2 s after firing (cont).
1.
To determine
the velocity of the projectile 2 s after firing, find both
the vertical and horizontal components of the projectile's velocity;
since velocity is a vector, both magnitude and direction must be
calculated.
x(m)
v
1
2
3
t(s)
Projectile Launched at Angle
from the Ground
d. Determine the velocity of the projectile 2 s after firing (cont).
1.
Since there is no acceleration in the x-direction, the final velocity in
the x-direction is equal to the initial velocity in the x-direction.
Find the x-component of the initial velocity.
Projectile Launched at Angle
from the Ground
d. Determine the velocity of the projectile 2 s after firing (cont).
To find the vertical component of the projectile's velocity, use
the first kinematics equation in the y-direction to solve for the
vertical component of the velocity of the projectile at the time of
t = 2 s.
Projectile Launched at Angle
from the Ground
d. Determine the velocity of the projectile 2 s after firing (cont).
Add the horizontal and vertical components of the velocity
together using vector addition to find the magnitude of the
velocity. The diagram illustrates how drawing tail to tip and
drawing the resultant creates a triangle. Use the Pythagorean
Theorem to solve for the magnitude of the velocity.
vx
v
vy
Projectile Launched at Angle
from the Ground
d. Determine the velocity of the projectile 2 s after firing (cont).
vx
v
vy
The velocity is 37 m/s at
6.2 degrees below the
horizontal.
Projectile Launched at Angle
from the Ground
e. Suppose a second projectile is launched but at an angle of 67
degrees above the horizontal with the same initial speed. Without
using equations, compare the following for the first and second
projectile launches:
i. maximum height
The projectile with the larger angle with respect to the ground is
going to go higher. The projectile fired at 67 degrees will go higher.
More of its velocity is in the y direction.
Projectile Launched at Angle
from the Ground
e. Suppose a second projectile is launched but at an angle of 67
degrees above the horizontal with the same initial speed. Without
using equations, compare the following for the first and second
projectile launches (cont):
ii. speed at the apex
The projectiles at the apex have stopped going upwards so the
motion is only in the horizontal direction, where there is no
acceleration. The initial horizontal velocity is the same as the
horizontal velocity at the apex.
The projectile fired at 67° will go higher which means that more of
its initial velocity goes into vertical motion than horizontal motion.
The projectile fired at 23° will be moving faster at the apex.
Projectile Launched at Angle
from the Ground
e. Suppose a second projectile is launched but at an angle of 67
degrees above the horizontal with the same initial speed. Without
using equations, compare the following for the first and second
projectile launches (cont):
iii. landing speed
Both projectiles are fired with the same initial velocity and
experience the same acceleration (a = -g).
Consider just one projectile. It's x velocity at any point in its
trajectory will stay constant, equaling the initial x velocity.
Projectile Launched at Angle
from the Ground
e. Suppose a second projectile is launched but at an angle of 67
degrees above the horizontal with the same initial speed. Without
using equations, compare the following for the first and second
projectile launches (cont):
iii. landing speed
Consider the y velocity. When an object is launched into the air,
its initial y velocity decreases due to the negative gravitational
acceleration until it reaches the apex of its orbit. The velocity
then increases as it heads towards the ground - subject to the
same acceleration.
The time and distance of the upward and downward motion are
the same. So, the y velocity right before it hits the ground is the
same as its initial y velocity.
Projectile Launched at Angle
from the Ground
e. Suppose a second projectile is launched but at an angle of 67
degrees above the horizontal with the same initial speed. Without
using equations, compare the following for the first and second
projectile launches (cont):
iii. landing speed
We did not talk about the angle of the launch. It doesn't matter
for determining the final x and y velocities.
If the final x and y velocities are the same as the initial, then the
total final velocity and total initial velocity are the same.
Since the launch angle doesn't matter, the two launches have the
same landing speed.
Dropped Package
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Dropped Package
An air plane flies at a constant horizontal speed of 185 m/s. When
the air plane is 1420 m above the ground level, a pilot drops a
package of relief supplies for hikers isolated by a snow storm.
a. How long it will take the package to reach the ground?
b. How far horizontally will the package fly before it strikes the
ground?
c. What is the velocity of the package just before it strikes the
ground?
d. Compare the magnitudes of the velocity of the package and
the velocity of an object that is dropped from a mountain of height
1420 m when they strike the ground.
Dropped Package
y0 = 1420 m
v0x = 185 m/s
a. How long it will take the package to reach the ground?
Use the Second Kinematics equation in the y direction.
Use the positive root to match the
physical configuration.
Dropped Package
y0 = 1420 m
v0x = 185 m/s
b. How far horizontally will the package fly before it strikes the
ground?
Use the Second Kinematics equation in the x direction with the
value of time found in part (a).
Dropped Package
y0 = 1420 m
v0x = 185 m/s
c. What is the velocity of the package just before it strikes the
ground?
Find the x and y components of the velocity at t = 17.0 s, and use
the Pythagorean theorem and trigonometry to find the total
velocity.
with respect to the +x axis
Dropped Package
y0 = 1420 m
v0x = 185 m/s
d. Compare the magnitudes of the velocity of the package and the
velocity of an object that is dropped from a mountain of height
1150 m when they strike the ground.
The magnitude of the velocity of the package from the plane is
249 m/s. The magnitude of the velocity of a package dropped
from the mountain is the same as the y velocity of the plane
package, 167 m/s.
The difference is the contribution of the x velocity of the plane/
package on its total velocity.
Cannon Ball Flight
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Cannon Ball Flight
A rocket propelled cannon ball is fired from a specially equipped antique
cannon located at the edge of a 28.0 m tall cliff with an initial velocity of
540.0 m/s at an angle 48.0° above the horizontal.
a. How much time is required for the cannon ball to reach the
ground?
b. How far from the cliff will the cannon ball strike the ground?
c. What is the maximum height, above the ground, reached by the
cannon ball?
d. What is the landing velocity of the cannon ball (magnitude and
direction)?
e. Draw x(t), y(t), vx(t), vy(t), ax(t), and ay(t) for the motion of the
cannon ball.
Cannon Ball Flight
y = 28.0 m
v0 = 540.0 m/s at θ = 48.0°
a. How much time is required for the cannon ball to reach the
ground?
Resolve v0 into its x and y components.
Use the Second Kinematics equation in y to find time to ground.
Use the solve function on your
calculator, or one of the many
websites that calculate this, or do it
manually with the quadratic equation.
Use the positive root.
Cannon Ball Flight
y = 28.0 m
v0 = 540.0 m/s at θ = 48.0°
b. How far from the cliff will the cannon ball strike the ground?
Use the Second Kinematics equation in the x direction.
Cannon Ball Flight
y = 28.0 m
v0 = 540.0 m/s at θ = 48.0°
c. What is the maximum height, above the ground, reached by the
cannon ball?
Use the Third Kinematics equation in y to find the height above
the cliff, then add the height of the cliff above the ground.
correct significant figures (3)
Cannon Ball Flight
y = 28.0 m
v0 = 540.0 m/s at θ = 48.0°
d. What is the landing velocity of the cannon ball (magnitude and
direction)?
Use time from part (a).
with respect to the +x axis
Cannon Ball Flight
y = 28.0 m
v0 = 540.0 m/s at θ = 48.0°
e. Draw x(t), y(t), vx(t), vy(t), ax(t), and ay(t) for the motion of the
cannon ball.
Cannon Ball Flight
y = 28.0 m
v0 = 540.0 m/s at θ = 48.0°
e. Draw x(t), y(t), vx(t), vy(t), ax(t), and ay(t) for the motion of the
cannon ball (cont).
the y position starts 28 m above the ground, but that doesn't
show up in the scale of the position graph.
Fire Fighter and a Fire
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Fire Fighter and a Fire
A fire fighter is trying to shoot water straight into the middle of a house
window located at a height 6.00 m above the ground. The distance
between the fire fighter and the house is 8.00 m and she holds the fire
hose 1.80 m above the ground. The water leaves the hose with a
constant speed of 12.5 m/s. Initially, the fire fighter aims the hose at
61.0 ◌̊ above the horizontal and misses the window.
a. How much time it will take for the water flow to reach the house?
b. How far above the middle of the window does the water go?
c. What is the magnitude and angle of the velocity of water when it
strikes the house?
d. What must be the minimum angle and speed of the flow in order to
get water into the middle of the window?
Fire Fighter and a Fire
yfire hose = 1.80 m
xwindow = 8.00 m
v0water = 12.5 m/s at θ = 61.0°
ymiddle of window = 6.00 m
a. How much time it will take for the water
flow to reach the house?
Resolve v0 into its x and y components.
Then Second Kinematics equation in the x direction.
Fire Fighter and a Fire
yfire hose = 1.80 m
xwindow = 8.00 m
v0water = 12.5 m/s at θ = 61.0°
ymiddle of window = 6.00 m
b. How far above the middle of the window
does the water go?
Find the height of the water stream at t = 1.32 s (found in part (a))
using the Second Kinematics equation in y. Then subtract the
height of the middle of the window.
Fire Fighter and a Fire
yfire hose = 1.80 m
xwindow = 8.00 m
v0water = 12.5 m/s at θ = 61.0°
ymiddle of window = 6.00 m
c. What is the magnitude and angle of the
velocity of water when it strikes the house?
Fire Fighter and a Fire
yfire hose = 1.80 m
xwindow = 8.00 m
v0water = 12.5 m/s at θ = 61.0°
ymiddle of window = 6.00 m
d. What must be the minimum angle and
speed of the flow in order to get water into
the middle of the window?
The minimum angle is where the peak of the water's trajectory is
right at the middle of the window. A greater angle will result in the
water stream to be too high, a lesser angle will result in the water
stream hitting below the window.
Set vy = 0 and use the Third Kinematics equation to solve for v0y.
Fire Fighter and a Fire
yfire hose = 1.80 m
xwindow = 8.00 m
v0water = 12.5 m/s at θ = 61.0°
ymiddle of window = 6.00 m
d. What must be the minimum angle and
speed of the flow in order to get water into
the middle of the window (contd)?
Find the time for the water to reach the window using the First
Kinematics equation.
Find v0x using the Second Kinematics
equation and t = 0.931 s.
Fire Fighter and a Fire
yfire hose = 1.80 m
xwindow = 8.00 m
v0water = 12.5 m/s at θ = 61.0°
ymiddle of window = 6.00 m
d. What must be the minimum angle and
speed of the flow in order to get water into
the middle of the window (contd)?
Solve for the initial velocity.
A Motor Boat Crossing a
River
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A Motor Boat Crossing a River
A person is on a boat capable of a maximum speed of 5.4 m/s, and
wishes to cross a river that is 1900 m wide by aiming the boat to a
point directly across from the starting point. The speed of the water
current is 1.6 m/s parallel to the shore.
a. What is the relative velocity of the boat with respect to the shore?
b. How much time will it take the boat to cross the river?
c. How far down the river will the boat land?
A Motor Boat Crossing a River
A person is on a boat capable of a maximum speed of 5.4 m/s,
and wishes to cross a river that is 1900 m wide by aiming the boat
to a point directly across from the starting point. The speed of the
water current is 1.6 m/s parallel to the shore.
a. What is the relative velocity of the boat with respect to the shore?
Given: VB = 5.4 m/s W = 1900 m Vc = 1.6 m/s
Make a sketch with the boat aimed directly across the
river. The boat has a speed of 5.4 m/s through the water,
but this is not its speed over the ground due to the current.
VB
A Motor Boat Crossing a River
a. What is the relative velocity of the boat with respect to the shore?
Given: VB = 5.4 m/s W = 1900 m Vc = 1.6 m/s
The current (VC) pushes the boat to the right. So, to find
the relative velocity (VR) with respect to the shore, the
vectors need to be added geometrically (tail to tip).
VC
VB
VR
A Motor Boat Crossing a River
a. What is the relative velocity of the boat with respect to the shore?
Given: VB = 5.4 m/s W = 1900 m Vc = 1.6 m/s
Since the three velocity vectors form a right triangle, the
Pythagorean theorem is used to find the magnitude of the
relative velocity.
VC
VB
VR
A Motor Boat Crossing a River
a. What is the relative velocity of the boat with respect to the shore?
Given: VB = 5.4 m/s W = 1900 m Vc = 1.6 m/s
The angle, θ, at which the boat deflects from the line
connecting the starting and end points of the journey is
found by trigonometry.
VC
VB
VR
o
A Motor Boat Crossing a River
b. How much time will it take the boat to cross the river?
Given: vB = 5.4 m/s W = 1900 m vc = 1.6 m/s
The width, W, of the river is covered by the boat's velocity
across the river, VB. The current velocity, VC, is
perpendicular to the width and doesn't affect the time it
takes for the boat to cross the river.
VC
W
VB
VR
A Motor Boat Crossing a River
c. How far down the river will the boat land?
Given: vB = 5.4 m/s W = 1900 m vc = 1.6 m/s
The distance, D, that the boat is pushed by the current
down the river is affected by only the current velocity, VC.
Use the time found in part (b) that it takes for the boat to
cross the river to the other side.
VC
W
VB
VR