Answer to Question #139373 in Physics for Nijean Lanterna

Question

 A ball is kicked horizontally from a 45 degrees slope hill with a velocity of 20 m/s as shown

in figure below. How long will it take for a ball to hit the ground at the bottom?

Answer 1

The horizontal distance covered by the ball is:

$$L = \dfrac{v_0^2\sin2\alpha}{g}$$

where $v_0 = 20m/s$ is the initial distance, $\alpha = 45\degree$ is the launch angle and $g = 9.81m/s^2$ is the gravitational acceleration. The horizontal speed of the ball is:

$$v_x = v_0\cos\alpha$$

Thus, the ball covers distance $L$ in time:

$$t = \dfrac{L}{v_x} = \dfrac{v_0^2\sin2\alpha}{gv_0\cos\alpha} = \dfrac{2v_0^2\sin\alpha\cos\alpha}{gv_0\cos\alpha} = \dfrac{2v_0\sin\alpha}{g}$$

Obtain:

$$t= \dfrac{2v_0\sin\alpha}{g} = = \dfrac{2\cdot 20\cdot \sin45\degree}{9.81} \approx 2.88 s$$

Answer. 2.88 s.