A juggler performs in a room whose ceiling is $3 m$ above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
What is the time taken by the ball to reach the ceiling?

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Solution

Using $S = v t - \frac{1}{2} a t^{2}$
Taking upward direction to be positive.
So, $S = 3 m$ $a = - g = - 10 m / s^{2}$
Final velocity of ball $v = 0$
So, $3 = 0 - \frac{1}{2} (- 10) t^{2}$
$⟹ t = 0.77 s$

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A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.What is the time taken by the ball to reach the ceiling?

Using S=vt−12at2Taking upward direction to be positive.So, S=3 m a=−g=−10 m/s2Final velocity of ball v=0So, 3=0−12(−10)t2⟹ t=0.77 s

A juggler performs in a room whose ceiling is $3 m$ above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
What is the time taken by the ball to reach the ceiling?

Using $S = v t - \frac{1}{2} a t^{2}$
Taking upward direction to be positive.
So, $S = 3 m$ $a = - g = - 10 m / s^{2}$
Final velocity of ball $v = 0$
So, $3 = 0 - \frac{1}{2} (- 10) t^{2}$
$⟹ t = 0.77 s$