phsyics help?

Question

A dive bomber has a velocity of 265 m/s at an angle below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.30 km. Find the angle.

Answer 1

Hint:
Solve equations
(265 m/s) * cosα * t = 3300 m
(265 m/s) * sinα * t + 0.5 * (9.81 m/s²) * t² = 2150 m
-