Question

A ball is projected at angle 60' from horizontal with speed 10m/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:

ans is √3/2

on max hight it's vertical velocity will be zero so from there it will be perpendicular to the image velocity

hope m crrt tell me if you need breif explanation

Answer 2

The velocity of the ball and its image will be perpendicular at time 0.37 seconds

Explanation:

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

$u_y=10\sin60^\circ$

$u_x=10\cos60^\circ$

There is a horizontal mirror on the ground.

The path of ball and their image is symmetrical about mirror (horizontal)

The angle between ball and image is 90° at time t

So, angle between ball and mirror at time t is 45°

Therefore,

$\tan45^\circ=\dfrac{v_y}{v_x}$

$1=\dfrac{u_y-gt}{u_x}$

$10\cos60^\circ=10\sin60^\circ-9.8t$

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

$t=\dfrac{3.66}{9.8}\approx 0.37$

Hence, the velocity of the ball and its image will be perpendicular at time 0.37 seconds

#Learn more:

https://brainly.in/question/10875164