Question

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. \[\sqrt{8gH+4g ^{2}t ^{2}}\]

Answer 1

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.

Answer 2

tnx a lot linjaaho !!