## anonymous 4 years ago A football is kicked from ground level with an initial vertical velocity of 48 ft/s. How long is the ball in the air? a.3 s b.5 s c.8 s d.10s

1. anonymous

8 s

2. anonymous

thank you xishem for explaining it,

3. anonymous

Oh! Wait. The units don't match up... Let me fix that.

4. anonymous

$y=y_0+v_yt+\frac{1}{2}a_yt^2$$0=0+(48\frac{ft}{s})t+\frac{1}{2}(-9.81\frac{m}{s^2}t^2)$$0=t[48\frac{ft}{s}-4.905\frac{m}{s^2}t]$Set each factor to 0.$0=t$$t=0$And...$0=48\frac{ft}{s}-4.905\frac{m}{s^2}t$$4.905\frac{m}{s^2}t=48\frac{ft}{s}$$t=9.7859\frac{ft*s}{m}*(\frac{1m}{3.281ft})=3.0s$There we go! :P