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?
enter image source here

Ans : 30^@

1 Answer
Aug 30, 2017

drawn

  • 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^@