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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
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8 s
thank you xishem for explaining it,
Oh! Wait. The units don't match up... Let me fix that.
\[y=y_0+v_yt+\frac{1}{2}a_yt^2\]\[0=0+(48\frac{ft}{s})t+\frac{1}{2}(-9.81\frac{m}{s^2}t^2)\]\[0=t[48\frac{ft}{s}-4.905\frac{m}{s^2}t]\]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
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