Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval. The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15m \; at\; t = 2s.$ The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Answers (1)

Let $u_{1}$ be the speed of the 1^st ball, $u_{1} = 2u\; m/s$

& $u_{2}$ be the speed of the second ball, $u_{2} = u \; m/s$

Let $h_{2}$ be the height of the two balls before coming to rest & $h_{1}$ be the height covered by 1 ball before coming to rest.

Now, we know that,

$V^{2}=u^{2}+2gh$

$H=\frac{v^{2}}{2g}$

Thus, $h_{1}=\frac{u{_{1}}^{2}}{2g}$

$=\frac{4u^{2}}{2g}$

and $h_{2}=\frac{u^{2}}{2g}$

Now, $h_{1}-h_{2}=15\; \; \; \; \; \; \; \; ........(given)$

Thus, $u=10\; m/s$

$h_{1}=20\; m\; and\; h_{2}=5\; m$

Calculating time for 1^st ball,

$V_{1} = u_{1} + gt$

$0 = 20 - 10t_{1}$

Thus, $t_{1} = 2 s$

Now, calculating time for second ball,

$V_{2} = u_{2} + gt_{2}$

$0 = 10 - 10t_{2}$

This, $t_{2}=1\; s$

Thus, time intervals between these two balls will be,

$=t_{1} - t_{2}$
$=(2-1)$

$= 1\; second.$

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question