Question

If a projectile is shot at an angle of `(3pi)/8` and at a velocity of `7 m/s`, when will it reach its maximum height?

Answer 1

The projectile will reach the maximum height at time `t=(7/10).cos(pi/8)` seconds after being thrown.

Explanation:

The initial velocity of the projectile is inclined at an angle `theta` with the horizontal.
Therefore the vertical motion which can be treated independently from the horizontal motion...will reach a point where its final velocity is zero.

and that instant is the position of maximum height.

The initial velocity component in the vertical direction is `u * cos (pi/8)`.

So, we have

`v = u * cos(pi/8) - g * t`

where `v` is the final velocity, which is zero.

Therefore, the projectile will reach the maximum height at a

`t = {u * cos( pi/8) } /g`

Substituting the values of `u` and `g`, we have

`t = (7/10.) cos (pi/8)`