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A stone projected with a velocity u at a...

A stone projected with a velocity u at an angle (theta )with the horizontal reaches maximum heights `H_(1)`. When it is projected with velocity u at an angle `(pi/2-theta)` with the horizontal, it reaches maximum height `H_(2)`. The relations between the horizontal range R of the projectile, `H_(1) and H_(2)`, is

A

`R=4sqrt(H_1H_2)`

B

`R=4(H_1-H_2)`

C

`R=4(H_1+H_2)`

D

`R=(H_1^2)/(H_2^2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`H_1=(u^2sin^2theta)/(2g)` and `H_2=(u^2sin^2(90-theta))/(2g)=(u^2cos^2theta)/(2g)`
`H_1H_2=(u^2sin^2theta)/(2g)xx(u^2cos^2theta)/(2g)=((u^2sin2theta)^2)/(16g^2)=(R^2)/(16)`
`R=4sqrt(H_1H_2)`
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