Question

A body is projected vertically upward with speed 40 m/s.The distance by body in the last second of upward journey is [take g=9.8 m/s2 and neglect effect of air resistance ]
1)4.9 m
2)9.8 m
3)12.4 m
4)19.6 m
pls...help Me with this question and provide me answer with steps...pls

Answer 1

Total distance travelled by body = H
Distance travelled by body in last second = d

H = u^2 / (2g)
= 40^2 / (2 * 9.8)
= 1600 / 19.6 m
= 81.63 m

Time taken by body to reach height H is sqrt(2H/g)
= sqrt(2 * 81.63 / 9.8)
= 4.08 seconds

Use equation of motion...,
S = ut + 0.5at^2
H - d = 40*(t - 1) - 0.5g(t - 1)^2
81.63 - d = 40 * (4.08 - 1) - 0.5*9.8*(4.08 - 1)^2
81.63 - d = 76.71
d = 81.63 - 76.71
d = 4.9 m

First option is correct.

Answer 2

Distance covered in last second of upward motion will be equal to the distance covered in the first second of downward motion.
Detailed solution is in the attachment.

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