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The vertical range of the projectile is ...

The vertical range of the projectile is :

A

`a//b`

B

`sqrt(a)//2b`

C

`a^2//2b`

D

`a^2//4b`

Text Solution

Verified by Experts

The correct Answer is:
A
`y=ax-bx^(2))`
Range is solution of y=0 `rArr ax-bx^(2)=0 rArr x=a//b rArr" Range "=a//b`
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