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April 19, 2014

April 19, 2014

Posted by **Todd** on Wednesday, June 30, 2010 at 10:46pm.

What is the initial velocity and how long did it take for the arrow to reach 16.0 m above the launch point?

- Physics -
**drwls**, Wednesday, June 30, 2010 at 11:15pmVelocity (v) obeys the equation

v = Vo - gt = Vo - 9.8 t

where Vo is the initial velocity

At t=2 s

32 = Vo - 19.6 m

Therefore Vo = 51.6 m/s

y (the height) = Vo t - (g/2)t^2

= 51.6 t - 4.9 t^2

Set y = 16 and solve for the time t when that height is reached. Take the lowest of the two solutions of the quadratic equation; the other is the time coming back down.

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