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?

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Solution

The correct option is D
60

u = + 20 m/s ; a = - 10 m/s2 ; t = 6s

∴ s = ut + 1/2 $a t^{2}$

= 20(6) + 1/2 (- 10) $6^{2}$ = 120 - 5 × 36

= 120 - 180 = - 60m

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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 60u = + 20 m/s ; a = - 10 m/s2 ; t = 6s ∴ s = ut + 1/2 at2 = 20(6) + 1/2 (- 10) 62 = 120 - 5 × 36 = 120 - 180 = - 60m

The correct option is D
60

u = + 20 m/s ; a = - 10 m/s2 ; t = 6s

∴ s = ut + 1/2 $a t^{2}$

= 20(6) + 1/2 (- 10) $6^{2}$ = 120 - 5 × 36

= 120 - 180 = - 60m