projectile motion ( pls help me)

Question

An object is projected with a speed of 28.0 m/s at an angle of 60 degrees to the horizontal.

(b) Determine

1.   The time taken to reach the greatest height.

2.   The greatest height reached.

Answer 1

v_x = 28cos60° = 14 m/s; x = 14t m; v_y = 28sin60° - 9.8t = 24.25- 9.8t; y = 24.25t - 4.9t²;

1. At highest point v_y = 24.25 - 9.8t = 0; t = 24.25/9.8 = 2.47 s
2. The greatest height h = 24.25·2.47 - 4.9·2.47² = 30 m