Question

The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the ceiling of the hall?

Answer 1

For this question, we need to remember the formula of projectile motion for horizontal distance and height of the object. The height of the ceiling and the speed of the ball is mentioned in the question. Substituting the given values in the formula we can get answers.
Formula used:
Height of the projectile, $H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$
Range of the projectile, $R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$

Complete step-by-step solution:
According to the question, a ball is thrown in a hall such that it should not hit the ceiling of the hall. So first we need to calculate the maximum height that the ball can go
Given: Height (h) = 25m
Speed (u) = 40 m/s

Let us consider that the ball is thrown at an angle $\theta$ with horizontal.
We know the formula for calculating the maximum height the ball has attained-
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$
Where , $u$ is the speed of ball, $\theta$ is the angle of projection with horizontal and $g$ is the acceleration due to gravity having value $9.8m/{{s}^{2}}$.
Therefore $25=\dfrac{{{(40)}^{2}}{{\sin }^{2}}\theta }{2\times 9.8}$
Rearranging the terms $\begin{aligned} & {{\sin }^{2}}\theta =0.30625 \\\ & \sin \theta =0.5534 \\\ \end{aligned}$
The value of $\theta$ is found to be $\begin{aligned} & \theta ={{\sin }^{-1}}(0.5534) \\\ & \theta =33.60{}^\circ \\\ \end{aligned}$
So with angle $\theta =33.60{}^\circ$, the ball will hit the maximum height of the hall.
Now applying formula for horizontal range,

& R=\dfrac{{{u}^{2}}\sin 2\theta }{g} \\\ & =\dfrac{{{(40)}^{2}}\sin (2\times 33.60{}^\circ )}{9.8} \\\ & =\dfrac{1600\sin (67.2)}{9.8} \\\ & =150.53m \end{aligned}$$ Therefore the range is found to be 150.53m. Thus the ball can be thrown at a distance of 150.53m **Additional information:** The range of an object is the maximum horizontal distance the object can cover when projected at any angle. Similarly, maximum height is referred to as the greatest height the object projected can attain. **Note:** The unit of the parameters used should always be kept in mind while using formulas. Here the velocity is in $$m/s$$ and acceleration due to gravity in $$m/{{s}^{2}}$$. The formula for height and range looks alike as there is a minute difference therefore it should be used carefully.