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

v_y = dy/dt = 10- 2t

At maximum height, v_y = 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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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?.

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

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?.

y = 10t - t²

v_y = dy/dt = 10- 2t

At maximum height, v_y = 0m/s

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

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