Answer to Question #284934 in Physics for Rukawa

Question

A projectile is fired horizontally at a speed of 8 m/s from an 80-m-high cliff. Find the angle of impact

Answer 1

Let's first find the time that the projectile takes to reach the ground:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\times80\ m}{9.8\ \dfrac{m}{s}}}=4.04\ s.$$

The horizontal component of projectile's velocity remains the same during the flight. Let's find the vertical component of the projectile's velocity:

$$v_x=v_{0x}=8\ \dfrac{m}{s},$$ $$v_y=-gt=-9.8\ \dfrac{m}{s^2}\times4.04\ s=-39.6\ \dfrac{m}{s}.$$

Finally, we can find the impact angle from the geometry:

$$\theta=tan^{-1}(\dfrac{v_y}{v_x}),$$ $$\theta=tan^{-1}(\dfrac{-39.6\ \dfrac{m}{s}}{8\ \dfrac{m}{s}})=78.6^{\circ}\ below\ the \ horizontal.$$