Solutions to Physics I H Chapter 6 Homework Problems 1-5
1.) A tiger leaps horizontally from a 6.5-m-high rock with a speed of 3.5 m s . How far from the
base of the rock will she land?
Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps
from the rock. In the horizontal direction, vx 0  3.5 m s and ax  0 . In the vertical direction,
v y 0  0 , a y  9.80 m s , y0  0 , and the final location y  6.5 m . The time for the tiger to reach
2
the ground is found from applying Eq. 2-11b to the vertical motion.
y  y0  v y 0t  12 a y t 2

6.5m  0  0 
1
2
 9.8 m s  t
2
2

t
2  6.5m 
9.8 m s 2
 1.15 sec
The horizontal displacement is calculated from the constant horizontal velocity.
x  vx t   3.5 m s 1.15 sec   4.0 m
2.) A diver running 1.8 m s dives out horizontally from the edge of a vertical cliff and 3.0 s later
reaches the water below. How high was the cliff, and how far from its base did the diver hit the
water?
Choose downward to be the positive y direction. The origin will be at the point where the diver dives
from the cliff. In the horizontal direction, vx 0  1.8 m s and ax  0 . In the vertical direction,
v y 0  0 , a y  9.80 m s , y0  0 , and the time of flight is t  3.0 s . The height of the cliff is found
2
from applying Eq. 2-11b to the vertical motion.
y  y0  v y 0t  12 a y t 2

 y  0  0  12 9.80 m s 2
 3.0s 
2
 44 m
The distance from the base of the cliff to where the diver hits the water is found from the
horizontal motion at constant velocity:
x  vx t  1.8 m s  3 s   5.4 m
Solutions to Physics I H Chapter 6 Homework Problems 1-5
3.) An athlete executing a long jump leaves the ground at a 28.0º angle and travels 7.80 m. (a)
What was the takeoff speed?
(a)
Use the “Level horizontal range” formula from Example 3-8.
R
v02 sin 2 0
g
 v0 
Rg
sin 2 0

 7.80 m   9.80 m

sin 2 28.0
o

s2
  9.60 m s
4.) A ball thrown horizontally at 22.2 m s from the roof of a building lands 36.0 m from the base
of the building. How tall is the building?
Choose downward to be the positive y direction. The origin is the point where the ball is thrown from
2
the roof of the building. In the vertical direction, v y 0  0 , y0  0 , and a y  9.80 m s . The initial
horizontal velocity is 22.2 m/s and the horizontal range is 36.0 m. The time of flight is found from
the horizontal motion at constant velocity.
x  vx t  t  x vx  36.0 m 22.2 m s  1.62 s
The vertical displacement, which is the height of the building, is found by applying Eq. 2-11b to the
vertical motion.
y  y0  vy 0t  12 a y t 2


y  0  0  12 9.80 m s 2
 1.62 s 
2
 12.9 m
5.) A football is kicked at ground level with a speed of 18.0 m s at an angle of 35.0º to the
horizontal. How much later does it hit the ground?
Choose the point at which the football is kicked the origin, and choose upward to be the positive y
direction. When the football reaches the ground again, the y displacement is 0. For the football,


vy 0  18.0sin 35.0o m s , a y  9.80 m s and the final y velocity will be the opposite of the
2
starting y velocity (reference problem 3-28). Use Eq. 2-11a to find the time of flight.
o
o
v y  v y 0  18.0 sin 35.0  m s  18.0 sin 35.0  m s
v y  v y 0  at  t 

 2.11 s
a
9.80 m s 2