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A ball is thrown up vertically and returns to the thrower after $6$ s. What is the maximum height reached by the ball?

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44.1 m

88.2 m

22.05 m

29.4 m

answer is A.

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Detailed Solution

The maximum height reached by the ball is

44.1 m

.
We know that the ball will take the same amount of time to come down as it takes to go up. This means it will travel upwards for

3 s

, and then downwards for

3

s.
Given data and assumptions:
For the first

t = 3

seconds.
Initial velocity,

= u

Final velocity,

v = 0 m / s

Maximum height to which the ball will rise,

= h

Acceleration due to gravity,

g = - 9.8 m / s^{2}

(as ball is moving upwards & the acceleration is working downward)
By the first equation of motion

v = u + gt

u = v - gt

u = 0 - (- 9.8) \times 3

u = 29.4 m / s

Now, using second equation of motion,

h = ut + \frac{1}{2} g t^{2}

h = (29.4) \times (3) + \frac{1}{2} \times (- 9.8) \times {(3)}^{2}

h = 44.1 m

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