Question

A ball is projected perpendicularly from an inclined plane of inclination $\theta$ from horizontal, with velocity ${}^{\prime }{u}^{\prime }$ as shown in figure. The time after which the projectile is making angle ${45}^{\circ }$ with the inclined plane is

• A. $\frac{u}{g\sin \theta }$
• B. $\frac{u}{g\cos \theta }$
• C. $\frac{u}{g(\sin \theta +\cos \theta )}$
• D. $\frac{u}{g(\sin \theta -\cos \theta )}$

Answer 1

The correct option is C $\frac{u}{g(\sin \theta +\cos \theta )}$


Taking $X\&Y$ axis along the inclined plane and perpendicular to the inclined plane respectively as shown in figure.
$a_x=+g\sin \theta ,a_y=-g\cos \theta ,u_y=+u,u_x=0$
$\Rightarrow$At anytime ${}^{\prime }{t}^{\prime }$ the component of velocity vector,
$v_x=u_x+a_{x}t=0+g\sin \theta \times t$
$\therefore v_x=g\sin \theta t...\left(i\right)$
$v_y=u_y+a_{y}t=+u-g\cos \theta \times t$
$v_y=u-g\cos \theta t...\left(ii\right)$
$\Rightarrow$At anytime $t$ the angle $\left(\alpha \right)$ made by velocity vector from the $X-$axis is given by:
$\therefore \tan \alpha =\frac{v_y}{v_x}$
$\tan \alpha =\frac{u-g\cos \theta t}{g\sin \theta t}$
$\tan {45}^{\circ }=\frac{u-g\cos \theta t}{g\sin \theta t}$
$\Rightarrow g\sin \theta t+g\cos \theta t=u$

$\therefore t=\frac{u}{g(\sin \theta +\cos \theta )}$