Question

A stone is dropped from a height of 45m what will be the distance travelled by it during the last one second of motion

Answer 1

Explanation:

the time taken by the stone to reach the ground=>

s=ut+1/2at^2

45=10t^2 [u=0 m/s and a=g=9.8m/s2 ≈ 10 m/s2]

45=5t^2

t^2=9

t=3 s

the time taken by the stone to reach the ground= 3 s

velocity of the stone after 2 seconds =>

v=u+at

v=10*2

v=20 m/s. [v will become u]

so, the distance traveled by the stone in last one second

will given by=>

s= ut+1/2at^2

s=20+5

s=25m

the distance travelled by it during the last one second of motion =25m

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