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 ſtake 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​

Answer 1

Let us take sign convention upward as negative and downward as positive.

u = – 40 m/s

g = 9.8 m/s²

v = 0 m/s

v = u + gt

0 = – 40 + 9.8 × t

t = 40 / 9.8

t = 4.081 sec ~ 4 sec

Distance travelled in last second

S4 – S3 = u ( t4 – t3 ) + 1/2 g ( t4² – t3² )

= – 40 ( 4 – 3 ) + 0.5 ( 10 ) ( 16 – 9 )

= – 40 + 5 × 7

= – 40 + 35

= – 5

Close answer ( 1 ) 4.9 m

I took g as 10 that is why my answer is 5 m. Distance negative here means in the upward direction