A small ball is projected up a smooth inclined plane with initial speed of 10m/s along the direction at $30^{\circ}$ $30 ^@$ 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.?

Question

What is the inclination of the inclined plane?

Ans : $30^{\circ}$ $30^@$

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Answer 1 · 2017-08-30T15:30:19

drawn

the velocity of projection of the ball on the inclined floor $u = 10 m / s$ $u=10m"/"s$
the angle of projection with the bottom edge of the slope $α = 30^{\circ}$ $alpha=30^@$
The effective acceleration due to gravity towards the bottom edge will be $g sin θ$ $gsintheta$ , considering the angle of inclination of the slope is $θ$ $theta$ and acceleration due to gravity $g = 10 m / s^{2}$ $g=10m"/"s^2$
The net displacement of the ball along the normal to the bottom edge of the inclined plane during its motion of $t = 2 sec$ $t=2 sec$ on the inclined floor is zero .
Hence by applying equation of kinematics we can write

$0 = u {sin 30}^{\circ} \times t - \frac{1}{2} g sin θ \times t^{2}$ $0=usin30^@xxt-1/2gsinthetaxxt^2$

$\Rightarrow 0 = 10 {sin 30}^{\circ} \times 2 - \frac{1}{2} \times 10 sin θ \times 2^{2}$ $=>0=10sin30^@xx2-1/2xx10sinthetaxx2^2$

$\Rightarrow sin θ = {sin 30}^{\circ}$ $=>sintheta=sin30^@$

$\Rightarrow θ = 30^{\circ}$ $=>theta=30^@$