Q.1 A ball is dropped from a height 15 m above the ground at t=0.0. Each time it
ID: 1655758 • Letter: Q
Question
Q.1 A ball is dropped from a height 15 m above the ground at t=0.0. Each time it bounces from the ground, its rebound speed is 60% of its impact speed. At the instant the first ball hits the ground, a second ball is released from the same place. Take g = 9.8 m/s2. At what height above the ground will they collide? (Take the ground as y=0.0 m)
Q.2 A and B can both walk at a speed 2.6 m/s and run at a speed 4.7 m/s. They start off together to go a distance 54 m. A walks for one half of the time it takes him to reach the distance, and runs for the remaining half the time. B, on the other hand, walks for half the distance, and runs for the remaining half the distance. Calculate how much sooner will A reach the distance as compared to B.
Q.3 A ball is projected vertically upward at time t = 0.00 s, from the ground and experiences negligible air resistance. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 17 m/s. The time when it is 7 m above the ground and moving downwards is (in seconds)
Explanation / Answer
a) impact speed of first ball = mgh = 1/2mv^2
v = sqroot ( 2gh) = 17.146 m/s apprx
rebound speed= 17.146 (0.6) =10.29 m/s apprx
let they meet at height h above the ground
1/2 ( gt^2) = ( 15-h)---for scond ball
4.9 t^2 = 15-h
t^2= ( 15-h)/4.9
h = 10.29 (t) - 1/2 g ( t)^2----for rebounding ball
h = 10.29 sqroot ( 15-h)/4.9 - 4.9 ( 15-h)/4.9
h = 4.647 sqroot ( 15-h) - 15 + h
15 = 4.647 sqroot ( 15-h)
squaring on both sides,
225= 21.594 ( 15-h)
10.419 = 15-h
h =4.58 m apprx
b)For A = let the first half of time = t
54 = 2.6 t + 5.7 t
t = (54/ 2.6 + 4.7)=54/ (7.3)
total time for A =14.794 seconds apprx
For B
t1= 27/2.6 + 27/4.7= 16.128 seconds apprx
A will reach = 16.128-14.794= 1.33 seconds apprx
C) max height = 17^2/ 19.6 =14.744 m
height covered while falling = 14.744 -7=7.45 m apprx
velocity = sqroot ( 2 g h) = sqroot ( 2x 9.8 x 7.45) =12.08 m/s apprx
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