Question

# A red ball is thrown down with an initial speed of 1.2 m/s from a height...

A red ball is thrown down with an initial speed of 1.2 m/s from a height of 25 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 23.8 m/s, from a height of 0.8 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/s^2.

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

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

3)What is the maximum height the blue ball reaches?

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

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

#### Homework Answers

Answer #1

for red ball

initial velocity vr1 = 1.2 m/s

initial position yr1 = 0

for blue ball

initial velocity vb1 = 23.8 m/s

initial position yb1 = 0.8 m

---------------

1)

vr2^2 - vr1^2 = 2*ay*(yr2 - yr1)

vr2^2 - 1.2^2 = -2*9.81*(0-25)

vr2 = 22.2 m/s

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

2)

yr2 - yr1 = vr1*t + (1/2)*ay*t^2

0 - 25 = 1.2*t - (1/2)*9.8*t^2

t = 2.38 s

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

3)

at maximum height final speed = 0

vb2 = 0

vb2^2 - vb1^2 = 28ay*(yb2 - yb1)

0^2 - 23.8^2 = -2*9.81*(yb2 - 0.8)

yb2 = 29.7 m

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

4)

yb3 - yb1 = vb1*(1.8-0.6) + (1/2)*ay*(1.8-0.6)^2

yb3 - 0.8 = 23.8*(1.8-0.6) - (1/2)*9.81*(1.8-0.6)^2

yb3 = 22.3 m

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

5)

after time t the ball are at same height

for red ball

yr = yr1 + vr1*t + (1/2)*ay*t^2

for blue ball

yb = yb1 + vb1*(t-0.6) + 91/20*ay*(t-0.6)^2

yb = yr

25 + 1.2*t - (1/2)*9.81*t^2 = 0.8 + 23.8*(t-0.6) - (1/2)*9.81*(t-0.6)^2

t = 1.413 s

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