Answer to Question #316301 in Physics for yaw

Question

A point particle is projected from the top of a hill 5m above sea level. It is projected with initial horizontal speed of 2m/s.

a. How far from the foot of the hill does it land?

b. How long does it take to hit the ground?

c. With what speed does it hit the ground?

d. With what velocity does it hit the ground?

Answer 1

a. How far from the foot of the hill does it land?

$$L=v_0\sqrt{2h/g}=2\sqrt{2*5/9.8}=2\:\rm m$$

b. How long does it take to hit the ground?

$$t=\sqrt{2h/g}=\sqrt{2*5/9.8}=1\:\rm s$$

c. With what speed does it hit the ground?

$$v=\sqrt{v_0^2+2gh}=\sqrt{2^2+2*9.8*5}=10\:\rm m/s$$

d. With what velocity does it hit the ground?

$$\theta=\tan^{-1}\frac{v_y}{v_x}=\tan^{-1}\frac{\sqrt{2gh}}{v_0}$$

$$\theta=\tan^{-1}\frac{\sqrt{2*9.8*5}}{2}=79^\circ$$

The velocity is 10 m/s directed 79 degrees below horizontal.