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Question

A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface

The horizontal and vertical displacement x and y vary with time t in seconds v as :

X = 10*3^1/2 t

Y = 10t-t^2

The maximum height by ball is?.

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Solution

y = 10t - t2

vy = dy/dt = 10- 2t

At maximum height, vy = 0m/s

Time taken to reach the maximum height, t = 5 seconds

Maximum height, y = 10(5) - (5)2 = 25 m


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