Question

(A). The ceiling of a hall is $25m$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40m{{s}^{-1}}$can go without hitting the ceiling?
(B). A cricketer can throw a ball to a maximum horizontal distance of $100m$. How much high above the ground can the cricketer throw the same ball?
(C). A stone tied to the end of a string $80cm$ long is whirled in a horizontal circle with a constant speed. If the stone makes 40 revolutions in $25s$, what is the magnitude and direction of the acceleration of the stone?

Answer 1

Hint: Range is the total distance covered along the horizontal axis and height is the maximum distance reached along the vertical axis. The ratio of range to height is one-fourth of $\tan \theta $($\theta $ is the angle at which the projectile is thrown). When $\theta ={{45}^{o}}$ , then the range and projectile are maximum. A centripetal force acts on the stone due to which it follows circular motion

Formulas used:
$\dfrac{1}{2}m{{v}^{2}}=mgh$
$\dfrac{H}{R}=\dfrac{1}{4}\tan \theta $
$R=\dfrac{{{u}^{2}}}{g}$
$H=\dfrac{{{u}^{2}}}{2g}$
$a={{\omega }^{2}}r$

Complete answer:
(A). the ball follows a projectile motion.
Velocity along horizontal axis= $v\cos \theta $
Velocity along the vertical axis= $v\sin \theta $
At the highest point, kinetic energy becomes zero and the potential energy is maximum.
Therefore,
$\dfrac{1}{2}m{{v}^{2}}=mgh$ ----(1)
The maximum kinetic energy at the lowest point is equal to maximum potential energy at the highest point.
Here, $v$ is maximum velocity
$m$ is the mass of the ball
$g$ is acceleration due to gravity
$h$ is maximum height
Solving eq (1), we get,
$\begin{align}
  & {{v}^{2}}=2gh \\
 & \Rightarrow {{(v\sin \theta )}^{2}}=2\times 10\times 25 \\
 & {{(40\sin \theta )}^{2}}=500 \\
\end{align}$
$\therefore \sin \theta =\dfrac{\sqrt{5}}{4}$ ----(2)
From eq (2),
$\tan \theta =\sqrt{\dfrac{5}{11}}$ --- (3)
We know that,
$\dfrac{H}{R}=\dfrac{1}{4}\tan \theta $ --- - (4)
Here, $H$ is the maximum height
$R$is the maximum range
Substituting, $H=25m$and$\tan \theta =\sqrt{\dfrac{5}{11}}$ from eq (3) in eq (4) we get,
$\begin{align}
  & \dfrac{25}{R}=\dfrac{1}{4}\sqrt{\dfrac{5}{11}} \\
 & \Rightarrow R=\dfrac{100\sqrt{11}}{\sqrt{5}} \\
\end{align}$
$\therefore R=20\sqrt{55m}$

The maximum horizontal distance that a ball can go is $20\sqrt{55m}$.

(B). Given, the cricketer throws the ball to a maximum horizontal distance of $100m$.
Since the range is maximum, angle is ${{45}^{o}}$
The formula for maximum range is-
$R=\dfrac{{{u}^{2}}}{g}$
Substituting the values we get,
$100=\dfrac{{{u}^{2}}}{10}$
$\Rightarrow 1000={{u}^{2}}$ --- - (2)
The formula for maximum height is given as,
$H=\dfrac{{{u}^{2}}}{2g}$
Therefore,
$\begin{align}
  & H=\dfrac{1000}{2\times 10} \\
 & \therefore H=50m \\
\end{align}$

Therefore, the maximum height the cricketer can throw the ball to is $50m$.

(C). Given, a stone is whirled in a horizontal circle, it makes 40 revolutions in $25s$.
Angular displacement covered, $\theta $= $2\pi \times 40=80\pi $
Initial angular velocity, $\omega $= $\dfrac{\theta }{t}=\dfrac{80\pi }{25}=\dfrac{16\pi }{5}\,rad\,{{s}^{-1}}$
The stone is moving with centripetal acceleration; therefore its acceleration is given by-
$a={{\omega }^{2}}r$ -- - (1)
Here,$a$ is the centripetal acceleration
$r$ is distance from axis of rotation
The distance from the axis of rotation of the stone is- $80cm=0.8m$
Substituting values in eq (1), we get,
$\begin{align}
  & a=\dfrac{16\pi }{5}\times \dfrac{16\pi }{5}\times 0.8 \\
 & \therefore a=8.2{{\pi }^{2}}m{{s}^{-2}} \\
\end{align}$

Therefore, the stone is accelerating at $8.2{{\pi }^{2}}m{{s}^{-2}}$. The direction of acceleration is directed towards the centre.

Note:
In a projectile motion, velocity is resolved in its horizontal and vertical components along the x-axis and y-axis respectively and the horizontal component of velocity remains constant. In circular motion, the direction of velocity is tangential to the path followed by the object. Centripetal force acts towards the centre.