A ball is thrown vertically upwards at time t = 0 s, and reaches a height of 10 m. At what time will the ball hit the ground?

t = 2 \sqrt{2} s

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t = 2 s

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t = 4 s

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t = \sqrt{2} s

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Solution

The correct option is A $t = 2 \sqrt{2} s$
The time taken to reach the maximum height is calculated using the first and second equations of motion. Then, this time is doubled to calculate the total time of the journey.

$v = u + a t 0 = u + (- 10) t u = + 10 t s = u t + \frac{1}{2} a t^{2} 10 = 10 t^{2} + \frac{1}{2} (- 10) t^{2} t = \sqrt{2} s T o t a l t i m e = 2 \sqrt{2} s$

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A ball is thrown vertically upwards at time t = 0 s, and reaches a height of 10 m. At what time will the ball hit the ground?

The correct option is A t=2√2sThe time taken to reach the maximum height is calculated using the first and second equations of motion. Then, this time is doubled to calculate the total time of the journey. v=u+at0=u+(−10)tu=+10ts=ut+12at210=10t2+12(−10)t2t=√2sTotaltime=2√2s