Answer to Question #177000 in Physics for Jordan Robinson

Question

A ball is thrown at 4.7 m/s from the top of a 12 m cliff at an angle of 25 degrees below the horizontal. Determine the speed of the ball when it reaches the ground

Answer 1

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

$$y=y_0-v_0tsin\theta-\dfrac{1}{2}gt^2,$$ $$0=12-4.7t\cdot\ sin25^{\circ}-4.9t^2,$$ $$4.9t^2+1.98t-12=0.$$

This quadratic equation has two roots: $t_1=1.37\ s$ and $t_2=-1.78\ s$. Since time can't be negative the correct answer is $t=1.37\ s$.

The horizontal component of the ball's velocity doesn't change during the flight and can be found as follows:

$$v_x=v_0cos\theta=4.7\ \dfrac{m}{s}\cdot cos25^{\circ}=4.26\ \dfrac{m}{s}.$$

The vertical component of the ball's velocity can be found as follows:

$$v_y=v_0sin\theta-gt,$$ $$v_y=4.7\ \dfrac{m}{s}\cdot sin25^{\circ}-9.8\ \dfrac{m}{s^2}\cdot1.37\ s=-11.44\ \dfrac{m}{s}.$$

Finally, we can find the speed of the ball when it reaches the ground from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(4.26\ \dfrac{m}{s})^2+(-11.44\ \dfrac{m}{s})^2}=12.2\ \dfrac{m}{s}.$$