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A hiker stands on the edge of a cliff 49...

A hiker stands on the edge of a cliff `490 m` above the ground and throwns a stone horiozontally with an initial speed of `15ms^(-1)` neglecting air resistance.The speed with which it hits the ground in `ms^(-1)` is `(g=9.8ms^(2))`

A

9.8

B

99

C

4.9

D

49

Text Solution

Verified by Experts

The correct Answer is:
B

`y=h =x tan theta -(gx^(2))/(2u^(2)cos^(2) theta)`,
`R=x+x+2h rArr x=(R )/(2)-h and R =(u^(2) sin 2 theta)/(g)`
using the above equations
`h=((R )/(2)-h) tan theta -(1)/(R ) ((R )/(2)-h)^(2) tan theta`
`rArr R=2h cot ((theta)/(2))`
putting `theta =60^(@), h =4m" then "R=8 sqrt(3) m`
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