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Question
A ball is thrown straight upward and rises to a maximum height of 21 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?
From conservation of energy you can get that the initial velocity v_launch is
v_launch = sq.root(2*g*H) , g=9.81 m/s^2, H=21 m.
Then take two equations of motion
v(t) = v_launch - g*t
h(t) = v_launch*t - g*t^2/2
and eliminate time from them. This reduces to
h(v) = (v_launch^2 - v^2)/(2*g)
Now put your condition v = (1/2)*v_launch in:
h = 3*v_launch^2/(8*g) = 3*(sq.root(2*g*H))^2/(8*g) =
= (3/4)*H = 16 m (or 15.75 m)
Good luck.
Daniel
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Daniel Mazur
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