Answer to Question #22257 in Mechanics | Relativity for Iggz
Let also v be the initial velocity of the first ball, andh1=40m be the initial height.
Then v1 is directed up.
Since both ball move with constant accelerationg=9.8m/s^2 directed down, we have that
y1(t) = h1 + v*t- g*t^2/2
y2(t) = h1, for t < 3s
= h1 -g*(t-3)^2 / 2, for t > 3s
Let s be the moment at which both balls hit the ground.
y1(T) = y2(T) =0.
y2(T) = 0
h1 - g*(T-3)^2/ 2 = 0
g*(T-3)^2 / 2 =h1
T = 3 +square_root(2*h1/g)
= 3 +square_root(2*40/9.8)
Substitute the value of s into the identity:
y1(T) = 0
h1 + v*T -g*T^2/2 = 0
v*T = g*T^2/2 -h1
v = g*T/2 -h1/T
=9.8*5.8571/2 - 40/5.8571
= 21.870 m/s
Answer: v = 21.870 m/s
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