Question

A small ball is projected up a smooth inclined plane with initial speed of 10m/s along the direction at `30 ^@` to the bottom edge of the slope. It returns to the edge after 2 sec. The ball is in contact with the inclined plane throughout the process.?

What is the inclination of the inclined plane?

Ans : `30^@`

Answer 1

• the velocity of projection of the ball on the inclined floor `u=10m"/"s`

• the angle of projection with the bottom edge of the slope `alpha=30^@`

• The effective acceleration due to gravity towards the bottom edge will be `gsintheta`, considering the angle of inclination of the slope is `theta` and acceleration due to gravity `g=10m"/"s^2`

• The net displacement of the ball along the normal to the bottom edge of the inclined plane during its motion of `t=2 sec` on the inclined floor is zero .

• Hence by applying equation of kinematics we can write

`0=usin30^@xxt-1/2gsinthetaxxt^2`

`=>0=10sin30^@xx2-1/2xx10sinthetaxx2^2`

`=>sintheta=sin30^@`

`=>theta=30^@`