Question

A stone is projected vertically upward with an initial velocity Vo. The distance travelled by it in time 1.5Vo/g.

Answer 1

Hey There!!

Here we are given that initial velocity of the stone is $v_{\circ}$.


Stone is thrown vertically upwards. Only acceleration is g downwards.


Let us first see at what time will it reach its maximum height. 


$v = u + at \\ \\ \implies 0 = v_{\circ}-gt \\ \\ \implies t = \frac{v_{\circ}}{g}$


Thus, the stone will reach it's maximum height in time $\frac{v_{\circ}}{g}$ which is less than $1.5\frac{v_{\circ}}{g}$

Thus, we have to also consider the time when the stone starts faling from maximum height.



Let us now see what the maximum height is:

$v^2=u^2+2as \\ \\ \implies 0 = v_{\circ}^2 - 2gh \\ \\ \implies h=\frac{v_{\circ}^2}{2g}$


Also, we see that:

$1.5\frac{v_{\circ}}{g} = \frac{v_{\circ}}{g} + \frac{v_{\circ}}{2g}$


Now, we need to consider what distance the stone free falls for the next $\frac{v_{\circ}}{2g}$ seconds.




$s = ut+\frac{1}{2}at^2 \\ \\ \\ \implies s = 0 + \frac{1}{2} \times g \times \frac{v_{\circ}^2}{4g^2} \\ \\ \\ \implies s = \frac{v_{\circ}^2}{8g}$

This is the distance for which the stone falls after attaining maximum height.


So, we can find total distance by adding this to maximum height:

$\textrm{Total Distance }= d = \frac{v_{\circ}^2}{2g}+\frac{v_{\circ}^2}{8g} \\ \\ \\ \implies \boxed{d=\frac{5v_{\circ}^2}{8g}}$



Hope it helps
Purva
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