Question

Paragraph
A small ball is thrown with initial speed u at an angle theta (with the horizontal) from a smooth horizontal surface. The ball
makes several impacts with the surface. The coefficient of restitution is e.
Q)The angle which the resultant velocity makes with horizontal, just after 4th impact (collision)
(1)tan-1(e4 tanø)
and other options,the answer is 1 but i need explanation,plz no spam ,
​

Answer 1

Answer:

eu2sin(p)

Explanation:

Coefficient of restitution is the part of the velocity with which the ball bounces up again in case of a bouncing. It depends on elasticity of the body.

Velocity of the body initially = u

Angle of the body bouncing with the horizontal = p

So, according to the formula,

Final velocity =v(\text {let})=u \times \sin (p)=v(let)=u×sin(p)

V-u=a tV−u=at

0-(e u \sin (p))=-g t0−(eusin(p))=−gt

\frac{-e u 2 \sin (p)}{-g}=t

−g

−eu2sin(p)

=t

So, time required is \frac{e u 2 \sin (p)}{g}

g

eu2sin(p)