Question

An inclined plane makes an angle $\theta ={45}^{\circ }$ with horizontal. A stone is projected normally from the inclined plane, with speed $um/s$ at $t=0$ (as shown in the figure). x and y-axis are drawn from point of projection along and normal to inclined plane as shown. The length of incline is sufficient for the stone to land on it and neglect air friction.

Answer 1

An inclined plane makes an angle$\theta =45°$

With horizontal

A stone is projected normally

from the inclined plane

The speed $=um/s$

at $t=0$

$x-y$ are drawn from point of projection along and normal to inclined $l^2=a^2+b^2$

where

$\sin 45°=\frac{a}{l}l=\frac{a}{1/\sqrt{2}}=\sqrt{2}al=\sqrt{2}u$

$\left(A\right)$ The instant of time at which velocity of stone is paralle4l to x-axis$=\frac{\sqrt{2}u}{g}\to \left(3\right)sec=45°=\sqrt{2}$

$\left(B\right)$ The instant of time at which velocity of stone makes an angle $\theta =45°$ with positive x-axis in clockwise direction

$\frac{u}{\sqrt{2}g}\to \left(4\right)\sin 45°=\frac{1}{\sqrt{2}}$

$\left(C\right)$ The instant of time which (starting from $t=0$) components of displacement along x-axis becomes the range on inclined plane.

$\frac{2u}{g}t=0\sin 45°=2$

$\left(D\right)$ Time of light on inclined plane $=\frac{2\sqrt{2}u}{g}\sqrt{2}=sec45°$