First slide

Rectilinear Motion

Question

A stone is allowed to fall from the top of a tower 300m height and at the same time another stone is projected vertically up from the ground with a velocity 30m/s .Find when the two stones meet?

Easy

A

5s
B

10 s
C

15 s
D

20 s

Solution

Let ‘t’ be the time after which the two stones meet
$∴$ For a vertically projected upwards body $u = 30 m / s$
$\begin{array}{l} a = g = - 9.8 m / s^{2} \\ s = x t = t \end{array}$
$∴$ From Equation $s = u t + \frac{1}{2} a t^{2}$
$x = 30 t - \frac{1}{2} g t^{2} ........... (i)$
For a freely falling body
$\begin{array}{l} u = 0 \\ a = g = 9.8 m / s^{2} \\ s = y \\ t = t \\ ∴ From equation s = u t + \frac{1}{2} g t^{2} \end{array}$
$\begin{array}{l} y = \frac{1}{2} g t^{2} \dots \dots \dots . . (2) \\ Given that x + y = 300 \\ \Rightarrow 30 t = 300 \\ \Rightarrow t = 10 s \end{array}$

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