A red ball is thrown down with an initial speed of 1.0 m/s from a height of 26.0

Question

A red ball is thrown down with an initial speed of 1.0 m/s from a height of 26.0 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 23.5 m/s, from a height of 1.0 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? __________________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 1.9 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

Explanation / Answer

a)V=sqrt (1+2*9.8*26)=22.5964598m/s b)t=(22.5964598-1)/9.8=2.20372s c)h1=23.5*23.5/(2*9.8)=28.1760m 54.176 m above the wall 29.1760 m from the ground will reach by blue ball d)h=23.5*1.9-9.8*0.5*(1.9)^2=26.961 meter above the wall 52.961 m above the ground e)1*t+0.5*9.8*t^2=-23.5*(t-0.6)+0.5*9.8*(t-0.6)^2 t=(23.5*0.6+0.5*9.8*(0.6^2))/(24.5-0.5*9.8*1.2)=0.85198711063 sec

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