A ball is thrown up from the top of a tower at 20 m/s velocity. It took the ball 6 sec to fall to the bottom. How high is the tower?

20 m

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40 m

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30 m

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60 m

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Solution

The correct option is D
60 m

u = +20 m/s

a = -10 $m / s^{2}$

t = 6s

$∴ s = u t + \frac{1}{2} a t^{2}$

= $20 (6) + \frac{1}{2} (- 10) (6)^{2}$

= $120 - 5 \times 36$

= 120 - 180

= -60 cm

$∴$ Displacement is negative as it is in the negative if direction. Thus, height at the tower is 60 cm.

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A ball is thrown up from the top of a tower at 20 m/s velocity. It took the ball 6 sec to fall to the bottom. How high is the tower?

The correct option is D 60 m u = +20 m/s a = -10 m/s2 t = 6s ∴s=ut+12at2 = 20(6)+12(−10)(6)2 = 120−5×36 = 120 - 180 = -60 cm ∴Displacement is negative as it is in the negative if direction. Thus, height at the tower is 60 cm.