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Find the ratio of maximum horizontal ran...

Find the ratio of maximum horizontal range and the maximum height attained by the projectile. i.e. for ` theta_(0) = 45^(@)`

A

`a//b`

B

`sqrt(a)//2b`

C

`a^2//2b`

D

`a^2//4b`

Text Solution

Verified by Experts

The correct Answer is:
D
`y=ax-bx^(2)`
Maximum height is attained at solution of `(dy)/(dx)=0`
`rArr a-2bx=0 rArr x=a//2b rArr y=a(a/(2b))-b(a/(2b))^(2)=a^(2)/(4b)`
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