Answer to Question #140586 in Physics for RANZ

Question

A ball is thrown so that it just clears a 10-feet fence 60-feet away. If it left the hand 5 feet above the ground and at angle of 60 degrees to the horizontal, what was the initial velocity of the ball?

Answer 1

Alright, we have just what we need: 10-feet fence 60-feet away cleared by the ball. Look at it:

See this orange arrow? This the initial velocity. The horizontal distance (60 ft) is defined by the horizontal component of velocity and time of flight. The height (10-5=5 ft) is defined by the vertical component of velocity. Now it's time to express it mathematically. Assume v is the initial velocity (length of the arrow). Then the components will be

$$v_x=v\text{ cos}60°,\\ v_y=v\text{ sin}60°.$$

The height of the fence:

$$5=\frac{v_y^2}{2g},\\\space\\ v_y=\frac{\sqrt{10g}}{\text{sin}60°}=\frac{\sqrt{10\cdot32.17}}{\text{sin}60°}=20.7\text{ ft/s}\\\space\\$$