RECITATION
CHAPTER 3
3.12 A daring 510 N swimmer dives off a cliff with a running horizontal leap, as shown
in the figure below. What must her minimum speed be just as she leaves the top of the
cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below
the top of the cliff?
Let downward be the –y direction. a x  0 and a y   g.
To find the time corresponding to the displacement Δy = -9.00 m, we use the following
equation:
1
y  y0  v0 y t  a y t 2
2
This is a zero launch angle problem and hence the initial velocity in the y direction v0y is
zero.
t
2y

ay
2 9.00m 
 1.36s
 9.80m / s 2


The initial velocity v0  v0 x is calculated as follows:
Using the constant acceleration equation in the x direction,
1
x  x0  v0 x t  a x t 2
2
x
x  v0 x t or v0 x 
 v0
t
1.75m
v0 
 1.29m / s
1.36s
Note that the weight of the diver was not used to solve the problem.
1
3.13 A 10,000 N car comes to a bridge during a storm and finds the bridge washed out.
The 650 N driver must get to the other side, so he decides to try leaping it with his car.
The side the car is on is 21.3 m above the river, while the opposite side is a mere 1.80 m
above the river. The river itself is a raging torrent 61.0 m wide. (a) How fast should the
car be traveling just as it leaves the cliff in order to clear the river and land safely on the
opposite side? (b) What is the speed of the car just before it lands safely on the other
side?
Let –y be the downward direction. Also, we have a x  0 and a y   g.
(a) Use the information given for the y direction to find the time of flight.
1
y  y 0  v0 y t  a y t 2
2
2y
21.80m  21.3m 
t

 1.995s
ay
 9.80m / s 2
The constant acceleration equation for the x direction x  x0  v0 x t 
v0 x 
x 61.0m

 30.6m / s .
t
1.995s
1
a x t 2 gives
2
(b) Since the car has no acceleration in the x direction, the velocity along the x
direction is constant
v x  v0 x  30.6m / s .
The velocity in the y direction at t = 1.995 s is calculated as follows:
v y  v0 y  a y t


v y   gt   9.80m / s 2 1.995s   19.6m / s
The magnitude of velocity just before it lands is
v  v x2  v y2  36.3m / s
2
3.18 A balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A
person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally
perpendicular to the path of the descending balloon, and 4.00 s later this person sees the
rock strike the ground. (See the figure below). (a) How high was the balloon when the
rock was thrown out? (b) How far horizontally does the rock travel before it hits the
ground? (c) At the instant the rock hits the ground, how far is it from the basket?
Let downward be the +y direction.
(a) The height when the rock was thrown out is calculated as follows:
1
y  y 0  v0 y t  a y t 2
2
1
2
y  20.0m / s 4.00 s   9.80m / s 2 4.00 s   158m
2
(b) The displacement of the rock in the x direction at t = 4.00 s is:
x  v0 xt  15.0m / s 4.00s   60.0m
(c) The basket descends (20.0 m/s) (4.00 s) = 80.0 m in 4.00 s, or it is
(158m - 80m) = 78.0 m above the ground when the rock reaches
the ground.
d  x2  y2 
60.0m2  78.0m2
3
 98.4m
3.20. A man stands on the roof of a 15.0-m-tall building and throws a rock with a velocity
of magnitude 30.0 m/s at an angle of 33.0° above the horizontal. You can ignore air
resistance. Calculate (a) the maximum height above the roof reached by the rock, (b) the
magnitude of the velocity of the rock just before it strikes the ground, and (c) the
horizontal distance from the base of the building to the point where the rock strikes the
ground.
Let downward be the -y direction.
v0 x  v0 cos  30.0m / s cos 33.0  25.2m / s; v0 y  v0 sin   30.0m / s sin 33.0  16.3m / s.
a x  0 and a y  0.
(a) At the maximum height, the velocity in the y direction vy is zero:
Using, v y2  v02y  2a y y
y 
 v02y
2a y
 16.3m / s 

 13.6m
2  9.80m / s 2
2


(b) The velocity in the x direction is constant all the time,
v x  v x 0  25.2m / s .
At the ground, we have y = -15.0 m, v y2  v02y  2a y y gives
v y   v0 y  2a y ( y  y0 )
2
v y   (16.3 m / s) 2  2(9.80 m / s 2 )( 15.0 m)  23.7 m / s.
The magnitude of velocity of the rock just before it strikes the ground is
v  vx  v y 
2
2
25.2m / s 2  23.7m / s 2
 34.6 m / s.
(c) Use the vertical motion to find the time of flight:
v y  v0 y  a y t
v y  v0 y
 23.7 m / s  16.3 m / s
 4.08 s.
ay
 9.80 m / s 2
The horizontal distance from the base of the building to the point where the
rock strikes the ground is
t

x  v0 x t  25.2m / s 4.08s   103m .
4
3.28. Two archers shoot arrows in the same direction from the same place with the same
initial speeds but at different angles. One shoots at 45° above the horizontal, while the
other shoots at 60.0°. If the arrow launched at 45° lands 225 m from the archer, how far
apart are the two arrows when they land? (You can assume that the arrows start at
essentially ground level.)
Here, we need to calculate the range for both the arrows. To find the range we proceed as
follows:
Find the time t for the projectile to reach the maximum height. The time of flight t’ is
t’=2t. At the maximum height vy = 0.
v0 y
v0 y
and hence t '  2t  2
.
v y  v0 y  gt  0 OR t 
g
g
2v0 x v0 y 2v0 cos  0 v0 sin  0  2v02 cos  0 sin  0
Range d  v0 x t ' 


g
g
g
2
v sin 2 0
OR d  0
.
g
Here, g = 9.80 m/s2 and  0 is the angle at which the projectile is launched.
The arrow which is shot at 45° has a range of 225 m. Using this information, the initial
speed at which the arrow is shot can be calculated as follows:
v02 


gd
9.80m / s 2 225m 

 2205m 2 / s 2
sin 2
sin 90
OR v0  47m / s
Both the arrows are shot with the same initial speed. Hence we can use this speed to
calculate the range for the arrow shot at 60.0° above the horizontal.


v02 sin 2 0
2205m 2 / s 2 sin 120
d

 195m.
g
9.80m / s 2
The distance between the two arrows when they land is 225 m – 195 m = 30 m.
3.30. Martian Olympics. The world record for the discus throw is 74.08 m, set by
Jürgen Schult in 1986. If he had been competing not on earth, but on Mars, where the
acceleration due to gravity is 0.379 what it is on earth, and if he had thrown the discus in
exactly the same way as on earth, what would be his Martian record for this throw?
Assume that the discus is released essentially at ground level.
From problem 28, the equation for range is
v 2 sin 20
d 0
.
g
It is given that g Mars  0.379 g Earth .
5
The discus is thrown with the same initial speed and direction on Mars as it was thrown
74.08mg earth
d
g
on earth. Hence, d Earth g earth  d Mars g Mars or d Mars  Earth earth 
 195m.
g Mars
0.379 g earth
3-62. During a testing program, the path of a projectile is recorded (see the figure below).
Use the information shown in the figure to find (a) the initial speed of the projectile, (b)
the angle at which the projectile is fired, and (c) the time during which the projectile is in
the air.
Figure gives the horizontal range to be R = 25.0 m and the maximum height to be
h = 4.90 m.
At the maximum height vy = 0. Using the constant acceleration equation we can get the
equation for maximum height h as follows:
v y2  v02y  2 gy  0
y  h 
v02y
2g

v0 sin  0 2
2g
(a) and (b)
 tan  0
h  v02 sin 2  0 
g
 2
 
 
R 
2g
4
 2v0 sin  0 cos  0 
4h
OR tan  0 
R
 4h 
 44.90m  
 0  tan 1    tan 1 
  38.1
 R
 25.0m 
Solving the equation h 

v0 sin  0 2
2g

for the initial speed v0 of the projectile
2 9.80m / s 2 4.90m 
 15.9m / s.
sin 38.1
x
R
25.0m


 2.00s
(c) The time of flight is t 
v0 x v0 cos  0 15.9m / s  cos 38.1
2 gh

gives v0 
sin  0
6
3-64. A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its
center, as shown in the figure below. The linear speed of a passenger on the rim is
constant and equal to 7.00 m/s. What are the magnitude and direction of the passenger’s
acceleration as she passes through (a) the lowest point in her circular motion and (b) the
highest point in her circular motion? (c) How much time does it take the Ferris wheel to
make one revolution?
(a) The magnitude of acceleration for circular motion is:
v 2 7.00m / s 

 3.50m / s 2 , upward .
R
14.0m
2
a rad 
.
v 2 7.00m / s 

 3.50m / s 2 , downward .
R
14.0m
2
(b) a rad 
(c) T 
2R 2 14.0m 

 12.6 s.
v
7.00m / s
7