Question

A ball is projected at an angle of 60 degrees above horizontal from a 3m above the ground. Find the maximum height that that ball went, total time of flight and the range. speed is 28m/s

Answer 1

The situation is as shown in the figure :

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Initial horizontal velocity

Initial vertical velocity

Net displacement = -3m

g=9.8 m/s^2

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a) Calculation of Max H :

At maximum height, the vertical velocity of the ball is 0,

0 = 24.2487^2 - 2 x 9.8 x H

=> H= 30m above the hill top

So the maximum height with respect to ground is 33m

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b) Calculation of Time of flight :

Net vertical displacement = s=-3m

g=-9.8m/s^2

s=ut+0.5 g t^2

=> -3 =24.2487t - 0.5x9.8xt^2

=> t=5 s

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c) The range = Time of flight x Ux = 14 x 5=70m