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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
Maximum heght, `H = (u^(2) sin^(2) theta)/(2g)`
and Time of flight, `T = 2(2u sin theta)/(g)`
So, `(H)/(T^(2)) = (u^(2)sin^(2)theta//2g)/(4u^(2)sin^(2)theta//g^(2)) = (g)/(8) = (5)/(4)`
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DC PANDEY-MOTION-Check Point 4.2
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