First slide

Projectile motion

Question

A gun kept on a straight horizontal road is used to shoot at a car travelling on the same road away from the gun at a uniform speed of $10 \sqrt{2} m s^{- 1}$ . The car is at a distance of 150 m from the gun when it is fired at an angle of 45° to the horizontal. With what speed should the shell be projected so that it hits the car? Take g =10 ms^-2.

Moderate

A

20 ms^-1
B

30 ms^-1
C

40 ms^-1
D

50 ms^-1

Solution

Let at t = 0, O and A are the positions of the gun and the car. Let us say that at time T, the shell and the car reach B simultaneously so that the shell hits the car when it is at a distance OB from the gun. Let u be the speed of projection of the shell. Then, initial horizontal component of velocity of the shell $= u \cos 45^{\circ} = \frac{u}{\sqrt{2}}$ and initial vertical component $= u \sin 45^{\circ} = u / \sqrt{2}$ . The car takes this time to cover the distance AB while the shell covers the distance OB in this time.
Now OB = OA + AB = 150 m + AB.
Distance AB is given by $AB = 10 \sqrt{2} \times \sqrt{2} \times \frac{u}{g} = 20 u / g$

And    $OB = \frac{u}{\sqrt{2}} \times \frac{\sqrt{2} u}{g} = u^{2} / g$
$∴$    $\frac{u^{2}}{g} = 150 + \frac{20 u}{g}$
or    $u^{2} - 20 u - 1500 = 0 (∵ g = 10 {ms}^{- 2})$
The positive root of this quadratic equation is u = 50 ms^-1.

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