Question

A cricket ball is thrown at a speed of 28m/s in a direction 30° above the horizontal. Calculate maximum height the time taken by the ball to return to the same level, and the horizontal distance from the thrower to the point where the ball returns to the same level.

Answer 1

maximum height = (V sin0)² / 2g 
= (28 sin30)²/ 2 x 9.8

= 14 x 14 / 2 x 9.8
= 10m

time taken to return to the same level
=2 v sin0/ g
= 2 x 28 x sin30/ 9.8
= 2 x 28 x 1/ 2 x 9.8 
= 2.9 sec

distance from thrower to where the ball returns
= v² sin 2Ф/ g
= 28 x 28 x sin60/ 9.8 
= 69.3 m

Answer 2

Given ,

Initial speed u = 28 m/s

Angle of projection $\alpha$ = 30°

acceleration due to gravity g = 9.8 m/s

To find :

1. max height gained by ball
2. time of flight
3. horizontal range of projectile .

Explanation :

The ball thrown has an initial velocity and is thrown at certain angle with respect to the horizontal hence , according to Projectile motion ,

1. Max. height gained by ball = $\frac{u^{2}sin^{2}\alpha }{2g}$

  substituting the values        = $\frac{(28)^{2} sin^{2}30 }{(2)(9.8)}$

                                                = 10 m

2. Time of flight  = $\frac{2u sin\alpha }{g}$

      substituting  = $\frac{(2)(28)sin 30}{9.8}$

                            = 2.86 s

3. Horizontal range of projectile = $\frac{u^{2} sin2\alpha }{g}$

                                substituting  = $\frac{28^{2} sin 2(30) }{9.8}$

                                                     = 68 m (approx)

Final answer :

Hence , the max. height , time of flight and the max range of the projectile are 10m , 2.8s and 68m approximately .