Question

# A body is projected vertically upward with speed 40 m/s. The distance travelled by body in the last second of upward journey is [ take g=9.8 m/s2 and neglect effect of air resistance]

A
4.9m
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B
9.8m
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C
12.4m
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D
19.6m
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Solution

## The correct option is A 4.9mStep 1: Calculate maximum height(H) [Ref. Fig. 1](Taking upward direction as positive)Let H be the maximum heightV= Velocity at top; V=0m/s; a=−9.8m/s2; s=H; u=40m/sSince acceleration is constant, therefore we can apply equation of motion V2−u2=2as ⇒02−402=2(−9.80)H ⇒H=81.63mStep 2: Calculate time required to reach H Applying equation of motion , s=ut+12at2 ⇒ 81.63=40t−12(9.80)t2 ⇒ t=4.08s Step 3: Calculate distance travelled by body in last second of upward journey [Ref. Fig. 2]Let the distance covered in last second =d ∴(H−d)m distance will be covered in (t−1) secondsApplying equation of motion, s=ut+12at2 ⇒(H−d)=40(4.08−1)−12×9.80(4.08−1)2 ⇒ 81.63−d=76.71 ⇒ d=4.91mHence option A is correct.

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