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The speed of a projectile at its highest...

The speed of a projectile at its highest point is `v_1` and at the point half the maximum height is `v_2`. If `v_1/v_2 = sqrt(2/5)`, then find the angle of projection.

A

`45^@`

B

`30^@`

C

`37^@`

D

`60^@`

Text Solution

Verified by Experts

The correct Answer is:
D
d. `v_1 = v_2 cos beta = u cos theta`
`0 = (v_2sinbeta)^2 - 2g(H//2)`

`rArr v_(1)^(2) = v_(2)^(2) - gH`
Also `v_1/v_2 = sqrt(2/5)`
From above `v_1 = sqrt((2gH)/3), v_2 = sqrt((5gH)/3)`
`H = (U^2sin^2theta)/(2g) rArr sin^2theta = (2gHcos^2theta)/(v_(1)^2)`
or `tan^2theta = (2gH)/(v_(1)^(2)) or tan^2theta = (2gHxx3)/(2gH)`
or `tan theta = sqrt(3) or theta = 60^@`.
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