A blue ball is thrown upward with an initial speed of 19.8 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 19.8 m/s, from a height of 0.8 meters above the ground. 2.4 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.6 m/s from a height of 21.9 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. What is the speed of the blue ball when it reaches its maximum height? How long does it take the blue ball to reach its maximum height? What is the maximum height the blue ball reaches? What is the height of the red ball 3.12 seconds after the blue ball is thrown? How long after the blue ball is thrown are the two balls in the air at the same height?

Explanation / Answer

1)
At maximum height, blue ball will be at rest
SO its velocity is 0
Answer: 0 m/s

2)
Vi = 19.8 m/s
Vf = 0 m/s
a = -9.81 m/s^2
t = ?
use:
Vf = Vi + a*t
0 = 19.8 + (-9.81)*t
t = 2.02 s
Answer: 2.02 s

3)
d= do + Vi*t + 0.5*a*t^2
=0.8 + 19.8*2.02 + 0.5* (-9.81)*(2.02)^2
=0.8 + 40 - 20
= 20.8 m
Answer: 20.8 m

4)
red ball has been travelling for 3.12 - 2.4 = 0.72 s
do = -21.9 m
Vi = 10.6 m/s
a = 9.8 m/s^2

d= do + Vi*t + 0.5*a*t^2
= -21.9 + 10.6*0.72 + 0.5*9.8*(0.72)^2
= -21.9 + 7.63 + 2.54
= -11.73 m
Answer: 11.73 m

I am allowed to answer only 4 parts at a time

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