Question

A cricket ball is hit with a velocity $25m{s^{ - 1}}$, ${60^0}$ above the horizontal. How far above the ground, the ball passes over a fielder $50\;m$ from the bat? Consider the ball is struck very close to the ground.
Take $\sqrt 3 = 1.7$ and $g = 10m{s^{ - 2}}$.
(A) $6.8\;m$
(B) $7\;m$
(C) $5\;m$
(D) $10\;m$

Answer 1

The cricket ball is undergoing a projectile motion. We know the velocity with which the ball is hit. The angle above the horizontal ground is also given in the question. We have to find out how far the ball will travel over a fielder over $50\;m$ from the bat.
Formula used:
$y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
where $y$ stands for the vertical position of the object, $x$ stands for the horizontal position of the object, $\tan \theta$ stands for the tangent of the launch angle, $g$ stands for the acceleration due to gravity, $u$ stands for the initial velocity, $\theta$ stands for the launch angle.

Complete Step by Step Solution:
The ball is hit with a velocity $25m{s^{ - 1}}$ .
The launch angle is ${60^0}$.
The fielder is at a distance of $50\;m$.
The equation of trajectory is given by,
$y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
The horizontal position, $x = 50m$
The launch angle, $\theta = {60^0}$
The acceleration due to gravity, $g = 10m/{s^2}$
The initial velocity, $u = 25m/s$
Substituting these values in the above equation,
$y = 50\tan 60 - \dfrac{{10 \times {{50}^2}}}{{2 \times {{25}^2}{{\cos }^2}60}}$
We know that $\tan {60^0} = \sqrt 3$ and $\cos {60^0} = \dfrac{1}{2}$.
Solving, we get
$y = 50\sqrt 3 - \dfrac{{10 \times 50 \times 50}}{{2 \times 25 \times 25 \times {{\left( {\dfrac{1}{2}} \right)}^2}}}$
$y = 86.60 - 80$
$y = 6.60 \approx 6.8m$
Therefore, the answer is

Option (A): $6.8\;m$

Additional Information: The time required for the projectile to reach the horizontal plane through the point of projection. The vertical height of the highest point on the trajectory from the point of projection is the maximum height of the projectile. To get maximum range the body should be projected at an angle of projection of ${45^0}$.

Note: An object projected into the air with a velocity is called a projectile. The projectile moves under the influence of the gravity of the earth. The path of the projectile is a parabola. The distance between the point where the trajectory meets the horizontal line and the point of projection through the point of projection is called the horizontal range of a projectile.