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.

• A.
  $160$
• B.
  $190$
• C. $175$
• D.
  $135$

Answer 1

The correct option is C $175$

Step 1: Draw diagram showing components.

Step 2: Find velocity when the particle collides with plane.

Given, projection velocity, $u=50m/s$
So, horizontal component of velocity,
$u_x=50\cos {53}^{\circ }=30m/s$
And acceleration along horizontal direction,
$a_x=0$
So horizontal velocity of the projectile remains constant throughout the motion,
Hence,
$\frac{v}{\sqrt{2}}=30$
$\Rightarrow v=30\sqrt{2}m/s$

Step 3: Find time taken by the projectile to reach the plane.

Formula used:$v=u+at$
Vertical component of initial velocity,
$u_y=50\sin {53}^{\circ }=40m/s$
And acceleration, $a_y=-g$
Let the time taken by the projectile to reach the incline plane be $t$,
So,
$v_y=u_y+a_{y}t$
$-30=40-10t$
$\Rightarrow t=7s$

Step 4: Find the distance between the point of projection and the foot of inclined plane.

The height of the incline plane where projectile hits it will be equal to the $y-$component of the projectile at $t=7s$.
So,
$h=u_{y}t-\frac{1}{2}g{t}^2$
$h=40\left(7\right)-\frac{1}{2}\times 10\times (7{)}^2$
$\Rightarrow h=35m$
As the angle of incline plane is given ${45}^{\circ }$. So, its height and horizontal distance will be same,
Hence, distance between the point of projection and foot inclined plane $=R-h$
$=u_{x}t-h$
$=\left(30\right)\left(7\right)-35=175m$

Final answer: (B)