Question

The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball can be thrown with a speed of $56m/s$ without hitting the ceiling of the hall is (take $g=9.8m/s^2$)
A) $25^\circ$
B) $30^\circ$
B) $45^\circ$
C) $60^\circ$

Answer 1

Hint: The above problem is based on the Projectile motion, where Maximum Horizontal distance is the maximum height up to which the object is projected. Formula for maximum horizontal range is:
${v^2}{\sin ^2}\theta = 2Hg$ (v is the velocity, H is the maximum height, g is the gravitational acceleration, $\theta $ is the angle at which the object is projected.
Using the above formula we will solve the given problem.

Complete step by step solution:
Let us define the maximum horizontal height and maximum horizontal range in more detail.
Maximum height: It is the maximum height to which a projectile rises above the horizontal plane of projection.
Maximum Horizontal Range: It is the horizontal distance from the point of projection to the point of projection to the point where the projectile comes back to the plane of projection. Its formula is given as:
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$ (R is the maximum range)
Now we will do the calculation part of the problem.
By using the formula which we have mentioned in the hint;
$ \Rightarrow {v^2}{\sin ^2}\theta = 2Hg$ (We will substitute the values of all the parameters)
$ \Rightarrow {56^2}{\sin ^2}\theta = 2 \times 40 \times 10$ (all the values have been substituted)
$ \Rightarrow {\sin ^2}\theta = \dfrac{{800}}{{{{56}^2}}}$ (arranged the terms)
$ \Rightarrow \sin \theta = .5$
$ \Rightarrow \theta = {\sin ^{ - 1}}(1/2)$ ($sin^{-1} (1/2) = 30^\circ$)
$\theta = {30^\circ}$
Angle at which the ball is thrown is $30^\circ$.
Maximum Horizontal range is calculated as;
We will divide the formula for horizontal range and maximum height;
$ \Rightarrow \dfrac{H}{R} = \dfrac{{\dfrac{{{v^2}{{\sin }^2}\theta }}{{2g}}}}{{\dfrac{{{v^2}\sin 2\theta }}{g}}}$ (cancelling the common terms)
$ \Rightarrow \dfrac{H}{R} = \dfrac{1}{4}\tan \theta $ .........................(1)
Now we substitute the values of all the parameters in equation (1)
$ \Rightarrow \dfrac{{40}}{R} = \dfrac{1}{4}\tan 30^\circ $ (We have substituted the values H and angle)
$ \Rightarrow R = \dfrac{{160}}{{\dfrac{1}{{\sqrt 3 }}}}$ (value of $tan \theta = \dfrac{1}{{\sqrt 3 }}$)
$ \Rightarrow R = 160\sqrt 3 $
Thus, will make an angle of $30^\circ$ without touching the ceiling at $160\sqrt 3 $ m range.

Option B is the correct option.

Note: Various examples of projectile motion are; A bullet fired from a gun, A bomb released from an airplane in level flight, An arrow released from bow, A javelin thrown by an athlete; While the motion of projectile is in the air, air acts as a resistance which we generally neglects.