Name: Motion With Variable Acceleration
Uploaded: 2019-07-19
Duration: 10 min 54 s
Description: You may be aware of the SUVAT equation, but do you think this can be applicable for all types of motion? No, what are those motion? Watch this to find out.

Question

A stone is projected from the point of a ground in such a direction so as to hit a bird on the top of a telegraph post of height and then attain the maximum height

2 h

above the ground. If at the instant of projection, the bird were to fly away horizontally with a uniform speed, find the ratio between the horizontal velocities of the bird and the stone, if the stone still hits the bird while descending.

Accepted Answer

Let $\vec{u} = u_{x} \hat{i} + u_{y} \hat{j}$ be the velocity of stone and $v$ is the velocity of the bird, then,

The maximum height that can be attained by the stone is $2 h$ , Hence, from equation of motion,
$2 h = \frac{u_{y}^{2}}{2 g}$

$u_{y} = 2 \sqrt{g h}$

Now, the bird is flying horizontally with velocity $v_{x}$ in time interval $0$ to $t_{2}$ . And the stone travels the same horizontal distance with velocity $u_{x}$ in time interval $t_{1}$ to $t_{2}$ , since, the horizontal component of projectile remains constant, so,

$(t_{2} - t_{1}) u_{x} = t_{2} v_{x}$

$\frac{v_{x}}{u_{x}} = \frac{t_{2} - t_{1}}{t_{2}} . . . (i)$

Taking motion along the vertical direction, applying the second equation of motion,

$h = u_{y} t - \frac{1}{2} g t^{2}$

$g t^{2} - 2 u_{y} t + h = 0$

$t_{1} = \frac{4 \sqrt{g h} - \sqrt{16 g h - 8 g h}}{2 g} = (2 - \sqrt{2}) \sqrt{\frac{h}{g}} . . . (ii)$

$t_{2} = (2 + \sqrt{2}) \sqrt{\frac{h}{g}} . . . (iii)$

Substituting $(ii)$ and $(iii)$ in $(i)$ , we get

$\frac{v_{x}}{u_{x}} = \frac{(2 + \sqrt{2} - 2 + \sqrt{2}) \sqrt{\frac{h}{g}}}{(2 + \sqrt{2}) \sqrt{\frac{h}{g}}}$

$\frac{v_{x}}{u_{x}} = \frac{2}{\sqrt{2} + 1}$ .

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