Question

Trajectory of particle in projectile motion is `y=(x-x^(2)//80)`, x and y are in metre.
Projectile range is on horizontal plane.

Answer 1

Correct Answer - `ararr r;brarrs; crarr p; drarrt`
`y=x tan theta-(gx^(2))/(2u^(2)cos^(2)theta)`
`y=(x-(x^(2))/(80)), theta=45^(@)`
`u^(2)=800`
`u=sqrt(800)`
`u_(y)=u_(y)-gt`
`u_(y)=20-10`
`u_(y)=10 , v_(x)=20`
`tan theta=(1)/(2) rArr theta=tan^(-1)((1)/(2))`