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A particle is projected with initial vel...

A particle is projected with initial velocity of `hati+2hatj`. The equation of trajectory is `(take g=10ms^(-2))`

A

`y=2x-15x^(2)`

B

`y=2x-25x^(2)`

C

`y=x-5x^(2)`

D

`y=2x-5x^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given, Horizontal component of initial velocity
`u_(x)=2=usin theta`
`tan theta=(u sin theta)/(u cos theta)=2/1=2`
The equation of trajectory of projectile motion is
`y=x tan theta-(gx^(2))/(2u^(2)cos^(2) theta)=xtan theta-(gx^(2))/(2(u_(x))^(2))`
`y=x xx2 -(10xx x^(2))/(2(1)^(2))=2x-5x^(2)`
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