Question

20 A ball is projected at angle 60° from horizontal with
speed 10 m/s. There is a horizontal mirror on the
ground. After how much time the velocity of the ball
will be perpendicular to the velocity of its image?​

Answer 1

Answer:

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Answer 2

The velocity of ball will be perpendicular to the velocity of it's image after 0.37 seconds.

Explanation:

A ball is projected at angle 60° from horizontal with  speed 10 m/s

u = 10 m/s

$\theta=60^\circ$

There is a horizontal mirror on the  ground.

• The path of image of ball and ball is symmetrical about horizontal mirror.

Therefore, the velocity vector of ball at any instant make equal angle with velocity of image of ball with mirror.

• If the angle between velocity of image and ball is 90° then angle between velocity vector at time t with mirror is 45°

$\tan\alpha=\dfrac{v_y}{v_x}$

$\alpha=45^\circ$

where, $\alpha=\text{ angle between velocity vector and horizontal}$

$v_y=\text{y-component of velocity vector at time t}$

$v_x=\text{x-component of velocity vector at time t}$

$\tan45^\circ=\dfrac{u\sin\theta-gt}{u\cos\theta}$

$\tan45^\circ=\dfrac{10\cdot \frac{\sqrt{3}}{2}-9.8\times t}{10\cdot \frac{1}{2}}$

$5=5\sqrt{3}-9.8t$

$t=\dfrac{5\sqrt{3}-5}{9.8}\approx 0.3735\ s$

Hence, the velocity of ball will be perpendicular to the velocity of it's image after 0.37 seconds.

Learn more:

Projectile motion

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