Question

A red ball is thrown down with an initial speed of 1.4 m/s from a height of 29 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 25.6 m/s, from a height of 1 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s².

1)

What is the speed of the red ball right before it hits the ground?

m/s

2)

How long does it take the red ball to reach the ground?

s

3)

What is the maximum height the blue ball reaches?

m

4)

What is the height of the blue ball 1.8 seconds after the red ball is thrown?

m

5)

How long after the red ball is thrown are the two balls in the air at the same height?

s

Answer 1

1)

for red ball

initial velocity v1iy = 14 m/s

initial position y1i = 29 m

before hittong ground

final position y1f = 0

final velocity v1fy = ?

v1fy^2 - v1iy^2 = 2*ay*(y1f-y1i)

v1fy^2 - 1.4 = -2*9.81*(0-29)

v1fy = 24 m/s

=======================

2)

y1f - y1i = v1iy + (1/2)*ay*t^2

0 - 29 = 1.4*t - (1/2)*9.81*t^2

time t = 2.58 s

==========================

3)

for blue ball

initial velocity v2iy = 25.6 m/s

initial position y2i = 1 m

at maximum height final velocity v2fy = 0

v2fy^2 - v2iy^2 = 2*ay*(y2f - y2i)

0 - 25.6^2 = -2*9.81*(y2f-1)

maximum height y2f = 34.4 m

===============================

4)

y2f - y2i = v2y*t + (1/2)*ay*t^2

time of travel for blue ball = 1.8 - 0.6 = 1.2 s

y2f - 1 = 25.6*1.2 - (1/2)*9.81*1.6^2

y2f = 19.2 m

================================

5)

height of red ball = height of blue ball

y1f = y2f

y1i + v1y*t + (1/2)*ay*t^2 = y2i + v2y*(t-0.6) + (1/2)*ay*(t-6)^2

29 + 1.4*t - (1/2)*9.81*t^2 = 1 + 25.6*(t-0.6) - (1/2)*9.81*(t-0.6)^2

t = 1.5 s