Ideal Projectile Motion
Any object that is released with some initial velocity and moves only under the influence
of gravity, neglecting air friction.
Characteristics of Projectile Motion.
Use the PHET Projectile
1.
2.
3.
4.
Some Equations to be used:
𝑉𝑜𝑥 = 𝑉𝑜 𝑐𝑜𝑠𝜃
𝑉𝑜𝑦 = 𝑉𝑜 𝑠𝑖𝑛𝜃
§
𝑅 = 𝑉𝑜𝑥 𝑡 = 𝑉𝑜 𝑐𝑜𝑠𝜃𝑡
1
1
𝑌 = 𝑉𝑜𝑦 𝑡 − 𝑔𝑡 2 = 𝑉𝑜 𝑠𝑖𝑛𝜃𝑡 − 𝑔𝑡 2
2
2
𝑉𝑦 = 𝑉𝑜𝑦 − 𝑔𝑡 = 𝑉𝑜 𝑠𝑖𝑛𝜃 − 𝑔𝑡
Note that vertical velocity (𝑉𝑦 ) 𝑖𝑠 𝑧𝑒𝑟𝑜 when maximum height is reached
𝑉𝑦 = 𝑉𝑜 𝑠𝑖𝑛𝜃 − 𝑔𝑡
𝑡=
𝑉𝑜 𝑠𝑖𝑛𝜃
𝑔
1
𝑌 = 𝑉𝑜𝑦 𝑡 − 𝑔𝑡 2
2
𝑉𝑜 𝑠𝑖𝑛𝜃
1 𝑉𝑜 𝑠𝑖𝑛𝜃 2
𝑉𝑜 2 𝑠𝑖𝑛2 𝜃
𝑌 = 𝑉𝑜 𝑠𝑖𝑛𝜃 (
)− 𝑔(
) =𝑌=
𝑔
2
𝑔
2𝑔
1
1
𝑌 = 𝑉𝑜 𝑠𝑖𝑛𝜃𝑡 − 2 𝑔𝑡 2 = 0 = 2 𝑔𝑡 2 = 𝑉𝑜 𝑠𝑖𝑛𝜃𝑡 = 𝑡 =
2𝑉𝑜 𝑠𝑖𝑛𝜃
𝑋 = 𝑉𝑜𝑥 𝑡 = 𝑉𝑜 𝑐𝑜𝑠𝜃 (
𝑔
)=
2𝑉𝑜2 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑔
=
2𝑉𝑜 𝑠𝑖𝑛𝜃
𝑔
𝑉𝑜 2 sin(2∗𝜃)
𝑔
Problem 1:
Two identical balls are shot with an initial velocity of 20.0 m/s One ball is shot at an angle of
15.0 degrees and another is shot an angle of 60.0 degrees. Find the range and the maximum
height.
For 15 degrees:
𝑉𝑜𝑥 = 𝑉𝑜 𝑐𝑜𝑠𝜃 = 20 𝑥 𝑐𝑜𝑠15 = 19.3 𝑚/𝑠
𝑉𝑜𝑦 = 𝑉𝑜 𝑠𝑖𝑛𝜃 = 20 𝑥 sin 15 = 5.2 𝑚/𝑠
𝑡=
𝑡=
𝑉𝑜 𝑠𝑖𝑛𝜃
𝑔
2𝑉𝑜 𝑠𝑖𝑛𝜃
𝑔
=
=
5.2
9.8
= 0.53 𝑠 (when it is at the maximum height)
2 𝑥 5.2
9.8
= 1.06 𝑠
𝑅 = 𝑉𝑜𝑥 𝑡 = 𝑉𝑜 𝑐𝑜𝑠𝜃𝑡 = 19.3 x 1.06 = 20.5 m
1
1
1
𝑌 = 𝑉𝑜𝑦 𝑡 − 𝑔𝑡 2 = 𝑉𝑜 𝑠𝑖𝑛𝜃𝑡 − 𝑔𝑡 2 = (5.2)(0.53) − (9.8)(0.53)2 = 1.38 𝑚
2
2
2
For 60 degrees:
𝑉𝑜𝑥 = 𝑉𝑜 𝑐𝑜𝑠𝜃 = 20 𝑥 𝑐𝑜𝑠60 = 10.0 𝑚/𝑠
𝑉𝑜𝑦 = 𝑉𝑜 𝑠𝑖𝑛𝜃 = 20 𝑥 sin 60 = 17.3 𝑚/𝑠
𝑡=
𝑡=
𝑉𝑜 𝑠𝑖𝑛𝜃
𝑔
2𝑉𝑜 𝑠𝑖𝑛𝜃
𝑔
=
=
17.3
9.8
= 1.77 𝑠 (when it is at the maximum height)
2 𝑥 17.3
9.8
= 3.53 𝑠
𝑅 = 𝑉𝑜𝑥 𝑡 = 𝑉𝑜 𝑐𝑜𝑠𝜃𝑡 = 10.0 x 3.53 = 35.3 m
1
1
1
𝑌 = 𝑉𝑜𝑦 𝑡 − 𝑔𝑡 2 = 𝑉𝑜 𝑠𝑖𝑛𝜃𝑡 − 𝑔𝑡 2 = (17.3)(1.77) − (9.8)(1.77)2 = 15.3 𝑚
2
2
2
Worksheet: Projectile Motion
1. A projectile is launched at 32.0 degrees to the horizontal with an initial speed of 25.0
m/s.
a. Find the horizontal and vertical components of initial velocity
b. Find the time it takes to reach the maximum height.
c. Find the time it takes to reach the end of trajectory.
d. Find the maximum height.
e. Find the maximum horizontal distance (range) reached.
2. A football is kicked at an angle of 37 degrees with an unknown initial velocity and lands
45.0 meters down field.
a. Find the maximum height of the ball
b. The time to reach the maximum height
c. The time in air before striking the ground.
3. A golf player hits a golf ball with an initial velocity of 83.0 m/s at 40 degrees. How long
did it take to reach the maximum height? How far did the ball travel?