Question

A ball thrown straight up into the air is found to be moving at 7.24 m/s after falling 2.27 m below its release point. Find the ball's initial speed. Thank you.

Answer 1

We know the equation of motion

..............................(1)

Where, v is the final velocity of the object

u is initial velocity of the object

a is acceleration of the object

S is total displacement of the object

Applying equation (1), between point B to C , we have

initial velocity = v_i = 0 m/s (at maximum height )

final velocity = 7.24 m/s (at point C)

a = g = 9.8 m/s² (acceleration due to gravity)

S = ( h + 2.27 ) m

So,   

or,  

or,  

Now applying equation (1) between the points A and B , we have

initial velocity = u (unknown)

final velocity = 0 m/s (at maximum height)

acceleration = -g = -9.8 m/s² (direction of acceleration is in opposite direction of the motion of the ball)

S = h = 0.4 m

So,

or,

or,

Hence, the initial velocity of the ball (at A) was 2.8 m/s.

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