# A bowling ball rolls off of a 2.00 meter tall table. It hits the ground 1.50 meters from the edge of the table.(Answer to 3 sig figs) What was the launch speed?_____ m/s What was the time in flight? _____ sec

Jan 12, 2017

It leaves the table traveling at 2.35 m/s and falls for 0.639 s.

#### Explanation:

This is a projectile motion problem. I will use the vertical motion to find the time in flight first, then I will calculate the launch speed. I think it is easiest this way.

Like any object that falls 2.00 m, the time required comes from

$\Delta y = {v}_{o} \Delta t + \frac{1}{2} a {\left(\Delta t\right)}^{2}$

In this case, ${v}_{o} = 0$, $\Delta y = - 2.00 m$ (because it is falling) and $a = - 9.8 \frac{m}{s} ^ 23$

$- 2.00 = - \frac{1}{2} \left(9.8\right) {\left(\Delta t\right)}^{2}$

${\left(\Delta t\right)}^{2} = \frac{2.00}{4.9} = 0.408$

$\Delta t = 0.639 s$

To travel 1.50 m in 0.639 s, the initial speed must have been

${v}_{o} = \frac{1.50 m}{0.639 s} = 2.35 \frac{m}{s}$