Physics... PLEASE HELP!!!

Question

A man stands on top of a 25 m tall cliff. The man throws a rock upwards at a 25 degree angle above the horizontal with an initial velocity of 7.2 m/s. What will be the vertical position of the rock at t = 2 seconds? What will be the horizontal position of the rock at that time?

Answer 1

Use the equation y= v_0yt - at²/2 +y₀ in this problem v_0y = 7.2sin25° = 3.04m/s and v_0x =7.2cos25° and y₀= 25m. So the equation becomes y= 3.04t -4.9t² + 25 and using t=2 gives y=11.48m which is the height above ground. The horizontal position is x= v_0xt which at t=2 gives x= 7.2cos25°•2 =6.53•2 = 13.06m.