Answer to Question #89604 in Classical Mechanics for sreelakshmi

Question

A ball is thrown from the top of a tower of height h with a velocity v. the ball strike the ground after time?

Answer 1

$\frac{dv}{dt} = -g \rightarrow v(t) = gt + v(0)$

$\frac{dx}{dt} = v(t) = gt + v(0) \rightarrow x(t) = x(0) + v(0)t - gt^2/2$

$0 = h - vt - gt^2/2$ - the moment when the ball strikes the ground, assuming that the initial velocity was directed downward (v(0) = -v)

$t = \frac{1}{g} \big(-v \pm \sqrt{v^2 +2gh} \big)$

The minus sign gives negative time, so we choose the plus sign

$$t = \frac{1}{g} \bigg(\sqrt{v^2 + 2gh} - v\bigg)$$