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For a projectile, the ratio of maximum h...

For a projectile, the ratio of maximum height reached to the square of flight time is `(g = 10 ms^(-2))`

A

`5:4`

B

`5:2`

C

`5:1`

D

`10:1`

Text Solution

Verified by Experts

The correct Answer is:
A
`H/T^(2) = ((u^(2) sin^(2)theta)/(2g))/((2u sin theta)/g)^(2) = (u^(2) sin^(2) theta)/(2g) xx g^(2)/(4u^(2) sin^(2) theta)`
`= 10/8 = 5/4`
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