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 (b) 9.8m (c) 12.4m (d) 19.6m

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.

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Answer 2

D is the answer of the questions