a cricket ball is thrown at a speed of 30 m\s in direction making an angle of 30 degree with the horizontal calculate maximum height, range of the ball, time taken by projectile to reach the maximum height?

 

Asked by nischithabby41 | 15th Jan, 2019, 06:10: PM

Expert Answer:

maximum height reached is calculated using the formula "v2 = u2 - 2gh " , where v is final velocity,
in this case zer when it reaches the highest point, u is initial speed, g is acceleration due to gravity and h is the height.
 
vertical component initial velocity = u × sin30 = 30×(1/2) = 15 m/s
 
h = u2 /(2g) = 15×15/(2×9.8) ≈ 11.5 m
 
time to reach maximum height  is calculated by " v = u - g t ", here final speed v =0 ,
u is initial vertical component of velocity 15 m/s
 
t = 15/9.8 = 1.53 s
 
Range of the ball = ( Horizontal component of velocity of projection )×time = ucosθ×2×1.53 = 30×(√3/2)×3.06 = 79.5 m

Answered by Ankit K | 15th Jan, 2019, 10:08: PM

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