Question

A ball is kicked near the edge of a cliff with a horizontal velocity of 10 m/s. 

What is the velocity upon impact if it hits the ground after 2 secs? 

What is the horizontal displacement of the ball? 

 

Answer 1

a) The horizontal component of ball's velocity doesn't chabge. The vertical component of the ball's velocity after 2 seconds will be:

$$v_y(t=2\ s)=v_0-gt=-9.8\ \dfrac{m}{s^2}\cdot2\ s=-19.6\ \dfrac{m}{s}.$$

We can find the impact velocity from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2},$$ $$v=\sqrt{(10\ \dfrac{m}{s})^2+(-19.6\ \dfrac{m}{s})^2}=22\ \dfrac{m}{s}.$$

b) The horizontal displacement of the ball can be found as follows:

$$x=v_0t=10\ \dfrac{m}{s}\cdot2\ s=20\ m.$$