A ball is thrown upwards from the top of a building with an initial velocity of 20 m/s and at an angle of 30 (degrees) with the horizontal. The height of the building from the ground level is 25m. Determine (i) where and when it will strike the ground (ii) velocity with which it strikes the ground (iii)maximum height reached by the ball above the ground level.
T total = (2Vi sin(theta))/2(9.81)
Range, R= (Vi)^2 sin2(theta)/ (9.81)
The Attempt at a Solution
i have solve for Vx= 20cos30 = 17.32 m/s
Vy=20sin30= 10 m/s