Question

A ball is projected from a point on the floor with a speed of $15\text{ m/s}$ at an angle of ${60}^{\circ }$ with the horizontal. Will it hit a vertical wall $5\text{ m}$ away from the point of projection and perpendicular to the plane of projection without hitting the floor? Will the answer differ if the wall is $\text{22 m}$ away?

Answer 1

Hint: In this question, a ball is projected, and we need to find out will it hit a vertical wall at a certain distance away from the point of projection. To solve this question, we need to find the range of the projectile. If the range of a projectile is greater than the distance of the wall from the point of projection, then the ball hits the wall else it will hit the floor first
Horizontal Range of a projectile, $\text{R = }{\displaystyle \frac{{\text{u}}^2\sin 2\theta }{g}}$
Where $\text{u}$ is the velocity of the projection
 $\theta$ is the angle of projectile with the horizontal
 $g=9.8\text{ m/}{\text{s}}^2$ (Acceleration due to gravity).

Complete answer:
We are given,
The velocity of the projection of ball, $\text{u}=15\text{ m/s}$
The angle of projection of the ball, $\theta ={60}^{\circ }$

We need to find out if the ball hits the vertical wall at a distance of $5\text{ m}$ away from the point of projection.
If the range of the projectile is greater than $5\text{ m}$ then the projectile hits the wall else, it does not hit the wall.
Thus, we need to find out the range of the projectile of the ball
Range, $\text{R = }{\displaystyle \frac{{\text{u}}^2\sin 2\theta }{g}}$
Now substituting the values in the formula we get,
 $R={\displaystyle \frac{{\left(15\right)}^2\sin {120}^{\circ }}{9.8}}$
Substituting $\sin {120}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$
 $\Rightarrow R={\displaystyle \frac{225\times \sqrt{3}}{2\times 9.8}}$
On solving we get,
 $R=19.88\text{ m}$
As the range $R$ is greater than $5\text{ m}$ the ball will definitely hit the vertical wall
For the second part of the question,
The wall is placed at a distance of $\text{22 m}$ from the point of projection
Since the range of a projectile of ball, $R=19.88\text{ m}$ is less than $\text{22 m}$ so the wall is not within the horizontal range of the ball
Hence, the ball will not hit the wall in the second case.

Note:
The horizontal range value is maximum if the angle of projection $\theta ={45}^{\circ }$ and for the same value of initial velocity horizontal range of a projectile for complementary angle is same. The formula $\text{R = }{\displaystyle \frac{{\text{u}}^2\sin 2\theta }{g}}$ is only valid for the case when projectile is projected obliquely on the surface of the earth.