Question

A cricket ball is thrown at a speed of $28m{{s}^{-1}}$ in a direction $30{}^\circ $ above horizontal. Calculate
a) the maximum height
b) the time taken by the ball in order to return to the same level.
c) The distance from the thrower to the point where the ball is returning to the similar level.

Answer 1

Hint: Here it is given the speed at which the ball is thrown and the angle at which the trajectory makes with the horizontal. The equations for the maximum height, time of flight, range of throw is to be used to calculate the same respectively for each of the questions. The answers of the same can be used to calculate the other quantities in the remaining questions.

Complete step by step answer:

Here it is given that,
Initial velocity is
$u=28m{{s}^{-1}}$
And the angle of projectile is
$\theta =30{}^\circ $
(a)The maximum height up to which a projectile can go is given by the formula,
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$
Where \[g\] the acceleration due to gravity.
Substituting the values in it,
\[H=\dfrac{{{28}^{2}}\times {{0.5}^{2}}}{2\times 9.8}=10m\]
Therefore the maximum height is calculated as
\[H=10m\]
(b) Time of flight of the cricket ball is given by the formula,
\[T=\dfrac{2u\sin \theta }{g}\]
Substituting the values in it,
\[\begin{align}
  & T=\dfrac{2\times 28\times \sin 30}{9.8} \\
 & T=2.86s \\
\end{align}\]
Therefore the time of flight is calculated as,
\[T=2.86s\]
(c) Range of the projectile is required to be calculate here,
The range of a projectile is given by the formula,
\[R=u\times \cos \theta \times T\]
Where \[T\] the time of flight of the projectile is.
Substituting the values in it will give,
\[\begin{align}
  & R=28\times \cos 30\times 2.86 \\
 & R=69.27m \\
\end{align}\]
Therefore the distance from the thrower to the point where ball is returning to the same level is given as,
\[R=69.27m\]

Note: A projectile is a body on which the only force acting is gravitational force. Gravity acts in order to make influence in the vertical motion of the projectile. This will cause a vertical acceleration. The horizontal motion of the projectile is caused due to the tendency of any object in motion to stay in motion at a specific velocity.