Question

A projectile is thrown with velocity of $50m/s$ towards an inclined plane from ground such that it strikes the inclined plane perpendicularly. The angle of projection of the projectile is $53^{\circ }$ with the horizontal and the inclined plane is inclined at an angle of $45^{\circ }$ to the horizontal. Find the distance (in $m$ ) between the point of projection and the foot of inclined plane.

Answer 1

Answer: 175

$u_x=50\cos 53^{\circ }=30m/s$,
$a_x=0,a_y=-g$
Now, $\frac{v}{\sqrt{2}}=30$
$\Rightarrow v=30\sqrt{2}m/s$
$v_y=u_y+a_{y}t$
$-30=40-10t$
$\Rightarrow t=7s$
$h=40(7)-\frac{1}{2}\times 10\times (7{)}^2$
$\Rightarrow h=35m$
Hence, distance between the point of projection and foot inclined plane $=175m$