The maximum height attained by a projectile when thrown at an angle `theta` with the horizontal is found to be half the horizontal range.Then `theta =`

Explanation

Let us consider an object thrown with initial velocity `vec(u)` an angle `theta` with the horizontal as in figure
`vec(u) = u_(z) hat(i) +u_(y) hat(j)`
The horizontal component of velocity `u_(x) = u cos theta` . The vertical component of velocity `u_(y) = u sin theta` .
Since the acceleration due to gravity is in the 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 gravitational force will push the projectile to move downward . It falls to the ground. there isno acceleration along the x direction throughout the motion. Hence, the horizontal component of the velocity `(u_(x)=u cos theta)` remains the same till the object reaches the ground .
Hence after the time t , the velocity along horizontal motion `v_(x) = u_(x) + a_(x) t` [ `:' u_(x) = u cos theta` ]
The horizontal distance travelled by projectile in time t is `s_(x) = u_(x) t + (1)/(2) a_(x) t^(2)`
Here, `s_(x) = x, u_(x)= u cos theta, a_(x)=0`
Thus , x ` = u cos theta t or t = (x)/(u cos theta ) " ". . . (1)`
For the vertical motion `v_(y) = u_(y) + a_(y) t`
Here `u_(y) = u sin theta, a_(y) = - g` (acceleration due to gravity acts opposite to the motion) . Thus
`v_(y) = u sin theta - g t " ". . . (2)`
the vertical distance travelled by the projectile in the same time t is
`s_(y) = u_(y) t + (1)/(2) a_(y) t^(2)`
Here , `s_(y) = y, u_(y) = u sin theta , a_(x) = - g. ` Then
`y = u sin theta t - (1)/(2) g t^(2)" ". . . (3)`
Substitute the value of t from equation (1) in equation (3) , we have

`y = u sin theta (x)/(u cos theta) - (1)/(2) g (x^(2))/(u^(2) cos^(2) theta)`
`y = x tan theta -(1)/(2) g (x^(2))/(u^(2) cos^(2) theta) " ". . . (4)`
Thus the path followed by the projectile is an inverted parabola.
Maximum height `(h_("max"))` :
The maximum vertical distance travelled by the projectile during its journey is called maximum height
For the vertical part of the motion,
`v_(y)^(2) = u_(y)^(2) + 2a_(y^(s))`
Here, `u_(y) = u sin theta, a = - g, s = h_("max")` , and at t he maximum height `v_(y) = 0`
`:. (0) = u^(2) sin ^(2) theta = 2 gh_("max")`
or `h_("max") = (u^(2) sin^(2) theta)/(2g)`
Horinontal range (R) : The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R) . It is found easily. Since the horizontal component of initial velocity remains the same.We can write `"Range R" = {{:("Horizontal component"),("of velocity" xx "time of flight"):}`
`= u cos theta xx T_(f)`
`R = u cos theta xx (2 u sin theta)/(g)`
` = (2 u^(2) sin theta cos theta)/(g)`
`rArr :. R = (u^(2) sin 2 theta)/(g)`