I would like to know the steps to take to solve please.
A golfer can hit a golf ball a horizontal distance of over 300m on a good drive. What maximum height will a 301.5m drive reach if it is launched at an angle of 25degrees to the ground?

Let v be the velocity the ball leaves the ground.

You can get the vertical and horizontal components of the velocity.

On the horizontal for displacement, you have:
s_h = u_ht

On the vertical for displacement, you have:
s_v = u_vt - \frac12 gt^2

1. Get t in terms of the horizontal distance and horizontal component of the velocity from the first equation.

2. Substitute this time divided by 2 into the second equation. (If it took 50 s to travel 300 m, the ball would be at its highest position at 25 s)

s_h = 301.5 m
g = 9.81 m/s^2

You can get s_v the maximum vertical displacement.

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