Question

A hiker stsnds on the edge of of a cliff 490m above the ground and throws a stone horizontally with a speed of 15m/s. the time taken by the stone to reach the ground is

Answer 1

Initial velocity(u)=0 m/s
displacement(s)=490 m
so,
s=ut+1/2 gt^2
=>490=0+1/2 x 9.8 x t^2
=>490/4.9=t^2
=>t=10 s
so,
now v=u+at
=>v=at=9.8x10=98 m/s

Answer 2

u=0 m/s

s=490 m

s=ut+1/2 gt^2

=>490=0+1/2 x 9.8 x t^2

=>490/4.9=t^2

=>t=10 s

so,

now v=u+at

=>v=at=9.8x10=98 m/s