At the highest point of oblique projection, which of the following is correct?

Explanation

This projectile motiontakes place when the intial velocity is not horizontal, but at some angle with the verticl , as shown in figure (Oblique projectiom).
Example,
Water ejected out of a hose pipe held obliquely.
Cannon fired in a battle ground,
consinder an object thrown with intial velocity at an angle `theta` with the horizontal.
then ` vecu=u_(x)hati+u_(y)hatJ`
Where `u_(x)=u costheta` is th horizontal componant and `u_(y)=u sintheta`, the vertical component of velocity.
since the accelereration due to gravity is in he direction opposite to te direction opposite to the direction of vertical component `u_(y)` this component will gradually reduce to zero at the maximum height of the projectile . At this maximum height the same garvitational force will oush the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion , So the horizonral component of the velocity `(u_(x)=u cos theta)` remains the same till the object reches the ground.

Hence after the time t, the velocity along horizontal motion ` v_(x)=u_(x)+a_(x)t=u_(x)=u cos theta`

Thuis , `x=u cos theta. t or t =(x)/(u costheta)`
Here `u_(y)=usin theta, a_(y)=-g` (acceleration due to gravity acts opposite to the motion).
Thus `v_(y)=u sin theta-g t`
The vertical distance travlled by the projectile in the same time t is
Here , `s_(y)=y, u_(y)=u sin theta, a_(x)=-g`
Substitute the value of t from equation (1) in equation (2), we have
`y=u sin theta (x)/(u costheta)-(1)/(2g)(x^(2))/(u^(2) cos^(2) theta)`
`y=xtantheta -(1)/(2g)(x^(2))/(u^(2) cos^(2) theta)`