Question

A marble rolls off of a table and travels a horizontal range of 0.35 m in the air before landing on the ground. If the table is 0.88 m tall, calculate the marble’s initial velocity as it leaves the table.

Answer 1

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

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot 0.88\ m}{9.8\ \dfrac{m}{s^2}}}=0.42\ s.$$

Finally, we can find the initial velocity as it leaves the table:

$$x=v_0t,$$ $$v_0=\dfrac{x}{t}=\dfrac{0.35\ m}{0.42\ s}=0.83\ \dfrac{m}{s}.$$

Answer:

$v_0=0.83\ \dfrac{m}{s}.$