Question #9846e

1 Answer
Steve J.
Sep 10, 2017

The distance from the base of the tower that body will be found is
v*sqrt((2*H)/(9.8 m/s^2))

Explanation:

The time to hit the ground is given by t in the equation
s = u*t + 1/2*a*t^2
where s = H, u = 0, a = g = (9.8 m)/s^2. Therefore
t = sqrt((2*H)/(9.8 m/s^2))

Let the velocity at which it was thrown be v. That means that in time t, the velocity v will have carried the body a horizontal distance
v*t = v*sqrt((2*H)/(9.8 m/s^2))

I hope this helps,
Steve

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