Question

A cricket ball is thrown at a speed of 56 m/s in a direction making an angle \[{{30}^{0}}\] with the horizontal. Calculate (a) maximum height, (b) Total time taken by the ball to return to the ground and (c) The distance from thrower to the point where the ball returns to the earth.

Answer 1

Hint:the ball is thrown making an angle with the horizontal. The path covered by the body is called projectile. Hence this is a problem of projectile motion and we have to use relevant formulas to find out the answers.

Complete step by step answer:
A-
Angle made with the horizontal is \[\theta ={{30}^{0}}\]
Initial velocity of the ball is u= 56 m/s
When the body is in projectile motion there is no acceleration in the horizontal direction and there is a constant vertical acceleration which is g.
Maximum height achieved by the projectile is given by \[H=\dfrac{{{u}^{2}}\sin \theta }{2g}\]
Substituting the values, we get,
\[H=\dfrac{{{56}^{2}}\sin 30}{2\times 9.8}=80m\]
So, the maximum height achieved by the projectile is 80m.

B-
Time taken by the ball to return to the ground is given by \[T=\dfrac{2u\sin \theta }{g}\]
Substituting the values, we get,
\[T=\dfrac{2\times 56\times \sin 30}{9.8}=5.7s\]
Thus, the total time taken by the body to return to the ground is 5.7 seconds.

C-
The distance from the thrower to the point where the ball returns to the earth is called Range. It is given by the formula
\[R=\dfrac{{{u}^{2}}sin2\theta }{g}\]
Substituting the values, we get, \[R=\dfrac{{{56}^{2}}\times sin60}{9.8}=277m\]
Thus the range is 277m.

Note:A projectile is any object thrown by the exertion of a force. Always in the formula the angle used is made with the vertical and if in the question it is given that angle is made with the vertical then we have to just subtract the given angle from \[{{90}^{0}}\].