A ball is thrown up from the top of a tower with 20 m/s velocity. It took the ball 6 sec to fall to the bottom. How much distance did the ball cover in the last second of its motion?

60 cm

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35 cm

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25 cm

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15 cm

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Solution

The correct option is B
35 cm

Last second is the 6th second

(Displacement at t = 6) - (displacement at t = 5) should give us the answer.

$= 20 (6) + \frac{1}{2} (- 10) (6)^{2} - [20 (5) + \frac{1}{2} (- 10) 5^{2}]$

= 120 - 180 - [100 - 125]

= - 60 - [-25]

= - 60 + 25

= - 35m

$∴$ Answer = 35m as we are asking for distance.

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A ball is thrown up from the top of a tower with 20 m/s velocity. It took the ball 6 sec to fall to the bottom. How much distance did the ball cover in the last second of its motion?

The correct option is B 35 cm Last second is the 6th second (Displacement at t = 6) - (displacement at t = 5) should give us the answer. =20(6)+12(−10)(6)2−[20(5)+12(−10)52] = 120 - 180 - [100 - 125] = - 60 - [-25] = - 60 + 25 = - 35m ∴Answer = 35m as we are asking for distance.