A blue ball is thrown upward with an initial speed of 20.8 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 20.8 m/s, from a height of 0.8 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.7 m/s from a height of 24.2 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.

so I have all but the final answer:

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

Hmax(blue)=2.12 s @ 22.848m

H(red @3.25s after blue ball is thrown)=16.42m

Explanation / Answer

for the blue ball:

u=20.8 m/s h= 0.8 m

after 2.5 s v=u+at

=> v = 20.8-9.81(2.5)

=-3.725 m/s => it travels down with a velocity of 3.725 m/s

height at that moment ,s = ut+(1/2)at^2

= 20.8(2.5)-0.5(9.81)(2.5^2)

= 21.34375 m

total height = 21.34475 +0.8 = 22.14475 m

at this moment red ball travels down with a velocity of 6.7m/s from a height of 24.2 m.

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