## anonymous 5 years ago a ball thrown vertically upwards from level ground is observed twice, at a height H above the ground, within a time interval t.The initial velocity of the ball was a. $\sqrt{8gH+g ^{2}t ^{2}}$ b. $\sqrt{8gH+\left( gT/2 \right)^{2}}$ c.$1/2\sqrt{8gH+g ^{2}t ^{2}}$ d. $\sqrt{8gH+4g ^{2}t ^{2}}$

1. anonymous

Here is one way to solve it. First we find the maximum height of the ball by the energy principle: $mgh_{\max}= \frac{1}{2}mv_0^2 \Leftrightarrow h_{\max}= \frac{v_0^2}{2g}$ Then let's denote the distance the ball has dropped from its highest point (after time t/2) with d, and therefore $d+H=h_{\max}\Leftrightarrow d=h_{max}-H$ Because the trajectory is symmetric, we can write $d=\frac{1}{2}g(\frac{t}{2})^2$ and $h_{max}-H=\frac{1}{2}g(\frac{t}{2})^2$ $\frac{v_0^2}{2g}-H=\frac{1}{2}g(\frac{t}{2})^2$ from which we can solve $v_0=\sqrt{\frac{1}{4}(gt)^2+2gH}=\frac{1}{2}\sqrt{(gt)^2+8gh}$ which is answer number c.

2. anonymous

tnx a lot linjaaho !!