Question

A projectile is thrown with velocity of $50 \,m / s$ 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

$u_{x}=50 \cos 53^{\circ}=30\, m / s$,
$ 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 $
$-30=40-10 t$
$\Rightarrow t=7\, s $
$h=40(7)-\frac{1}{2} \times 10 \times(7)^{2}$
$ \Rightarrow h=35 \,m$
Hence, distance between the point of projection and foot inclined plane $=175 \,m$