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Question

A ball thrown vertically upwards from the ground falls back to ground in 3s . Determine the velocity of projection of ball and maximum height attained by it.


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Solution

Step 1: Given data

The time of flight of the ball is 3s

The velocity of the ball at a maximum height, v=0ms-1.

Take the initial velocity as ums-1.

(Note: Consider the downward direction as negative throughout the motion of the ball and the acceleration due to the gravity of the earth as g=9.8ms-1)

Step 2: Find the initial velocity of projection of the ball

Time to reach the maximum height, t=timeofflight2=32s

As we know that the first equation of motion is, v=u+at

On substituting all the given data to the above equation we get,

⇒v=u-gt

⇒0=u-9.8×32

⇒u=14.7ms-1

Step 3: Determine the maximum height attained by the ball.

As we know the second equation of motion under gravity is S=ut+12at2

On putting the given data in the above equation we get,

⇒h=ut-12gt2

⇒h=14.7×32-12×9.8×322

⇒h=22.050-11.025

⇒h=11.025m

Thus, the maximum height attained by it is 11.025m.


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