Question

A projectile is fired with velocity at an angle of 60 to the horizontal

calculate the maximum height reached by the ball after 4s​

Answer 1

Let
`v_i` be the initial speed (m/s), so that
`v_(i,y)=v_i\sin60^\circ=\frac{\sqrt3}2v_i` (also m/s) is the vertical component of the initial velocity vector.

Recall that

`{v_f}^2-{v_i}^2=2a\Delta y`

where
`v_i` and
`v_f` are the initial and final velocities,
`a` is the acceleration, and
`\Delta y` is the vertical displacement. At its maximum height (which the question seems to say occurs after 4 seconds), the velocity is 0, and throughout its motion the projectile is under the influence of gravity so that
`a=-g(\rm m)/(\mathrm s^2)` and g = 9.80.

`-\left(\frac{\sqrt3}2v_i\right)^2=-2gy_(\rm max)\quad\quad(*)`

The projectile's height
`y` (m) at time
`t` (s) is

`y=v_(i,y)t-\frac g2t^2`

so that when t = 4 s, its height is

`y_(\rm max)=2\sqrt3v_i-8g\quad\quad(**)`

Solve
`(*)` and
`(**)` for
`v_i`, then solve for
`y_(\rm max)`:

`(*)\implies v_i=\sqrt{\frac{8gy_(\rm max)}3}`

`(**)\implies v_i=(y_(\rm max)+8g)/(2\sqrt3)`

Then

`\sqrt{\frac{8gy_(\rm max)}3}=(y_(\rm max)+8g)/(2\sqrt3)`

`\implies\frac{{y_(\rm max)}^2-16gy_(\rm max)+64g^2}4=0`

`\implies(y_(\rm max)-8g)^2=0`

`\implies y_(\rm max)=8g\approx\boxed{78.4\,\mathrm m}`