A ball thrown into the air from the ground reaches a maximum height of 50ft and travels 200ft horizontally before it drops again to the ground. Find the magnitude and direction of the velocity with which it was thrown.
Let the launch velocity be u and the angle be theta.
Maximum height reached, H = (u sin theta)^2 /2g
u^2 sin^2 theta = 2 * g * ...
Step-wise solution is provided.