Answer to Question #129966 in Physics for fi

Question

A marble rolling at a horizontal speed of 1.8ms_1 rolls off a bench with 1.2m height.
Determine the distance of the marble from the edge of the bench after it hits the floor for
the first time.

Answer 1

The equations of motion of the marble are $y(t) = H - \frac{g t^2}{2}, x(t) = v_0 t$, where $H = 1.2m$ is the height of the bench, $v_0 = 1.8 \frac{m}{s}$ is the initial horizontal speed of the marble, $g = 9.81 \frac{m}{s^2}$ is the gravitational acceleration.

When the marble reaches the ground, $y(t') = 0$, from where the time it takes to reach the ground is $t' = \sqrt{\frac{2 H}{g}}$ . Substituting this time into equation for x-coordinate, obtain the distance of the marble from the edge of the bench: $L = x(t') =v_0 t' = v_0 \sqrt{\frac{2 H}{g}} \approx 0.89 m$.