First slide

Rectilinear Motion

Question

If a ball is thrown vertically upwards with a speed ‘u’, the distance covered during the last ‘t’ seconds of its ascent is……..

Easy

A

$\frac{1}{2} g t^{2}$
B

$u t - \frac{1}{2} g t^{2}$
C

$u t$
D

$\frac{u t}{2 g}$

Solution

We know that the time of ascent $T = \frac{u}{g}$
Total distance covered by the ball in ‘T’ seconds
$\begin{array}{l} H = u T - \frac{1}{2} g T^{2} \\ = u (\frac{u}{g}) - \frac{1}{2} g {(\frac{u}{g})}^{2} \\ = \frac{u^{2}}{g} - \frac{u^{2}}{2 g} = \frac{u^{2}}{2 g} \dots \dots \dots . . (i) \end{array}$
Similarly distance covered during (T-t)sec
$\begin{array}{l} h = u [T - t) - \frac{1}{2} (T - t)^{2} \\ = u [\frac{u}{g} - t] - \frac{1}{2} g [{(\frac{u}{g} - t)}^{2}] \\ = (\frac{u^{2}}{g} - u t) - \frac{1}{2} g (\frac{u^{2}}{g^{2}} + t^{2} - \frac{2 u t}{g}) \\ = (\frac{u^{2}}{g} - u t) - (\frac{u^{2}}{2 g} + \frac{1}{2} g t^{2} - u t) \\ = \frac{u^{2}}{g} - u t - \frac{u^{2}}{2 g} - \frac{1}{2} g t^{2} + u t \\ = \frac{u^{2}}{2 g} - \frac{1}{2} g t^{2} \end{array}$
Distance covered during the last ‘t’sec of its ascent =H-h
$\begin{array}{l} = \frac{u^{2}}{2 g} - (\frac{u^{2}}{2 g} - \frac{1}{2} g t^{2}) \\ = \frac{1}{2} g t^{2} \end{array}$

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