Answer to Question #177410 in Mechanics | Relativity for Abbas

Question

A ball is thrown horizontally from the top of a building that is 26.01 m high. The ball strikes the ground at an angle 41.86 with respect to the horizontal. Neglecting air resistance, with what speed was the ball thrown?

Answer 1

$$v_y=v_x\tan{41.86}\\\sqrt{2gh}=v_x\tan{41.86}\\\sqrt{2(9.8)(26.01)}=v_x\tan{41.86}\\ v_x=25.2\frac{m}{s}$$