Question

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

Answer 1

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

`Deltay=v_o Deltat+1/2a (Deltat)^2`

In this case, `v_o=0`, `Deltay=-2.00m` (because it is falling) and `a=-9.8m/s^23`

`-2.00 = -1/2(9.8)(Deltat)^2`

`(Deltat)^2=(2.00)/(4.9)= 0.408`

`Deltat=0.639s`

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

`v_o=(1.50m)/(0.639s)=2.35 m/s`