A ball is thrown vertically upwards with a velocity of 20ms−1 from the top of a multistorey building. The height of the point where the ball is thrown 25 m from the ground. How long will it be before the ball hits the ground ? Take g=10ms−2. 

A ball is thrown vertically upwards with a velocity of 20ms−1 from the top of a multistorey building. The height of the point where the ball is thrown 25 m from the ground. How long will it be before the ball hits the ground ? Take g=10ms−2.

Answer:

Initial velocity of the ball = \large\bf 20ms^{-1}
The time it will take to reach the maximum height:
\large\bf v = u+gt
\large\bf \hspace{30pt}  0 = 20 + (-10)t
\large\bf \hspace{30pt}  t = 2s

The maximum height ball reached from building can be calculated by:
\large\bf s = ut+ \frac{1}{2} gt^{2}
\large\bf s = 20 \times 2 + \frac{1}{2} \times (-10) \times (2)^{2}
\large\bf s = 20 m

Maximum height from ground = \large\bf 20m + 25m
\hspace{30pt} \large\bf = 45m

Time the ball will take from maximum height to reach ground:
\large\bf s = ut+ \frac{1}{2} gt^{2}
\large\bf 45 = 0 \times t + \frac{1}{2} \times (10) \times (t)^{2}
\large\bf 45 = 5t^{2}
\large\bf t^{2} = 9
\large\bf t = 3s

So, the total time the ball will take to hit the ground = \large\bf 3s + 2s = 5s

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