Posted by **Dani** on Thursday, September 22, 2011 at 7:36pm.

An arrow is shot at a 30 degree angle with the horizontal. It has a velocity of 49m/s. The acceleration of gravity is 9.8 m/s2. How high will the arrow go?

- Physics -
**Steve**, Friday, September 23, 2011 at 12:38pm
Since we don't care about the horizontal motion, just ignore it. The initial vertical velocity

v0 = 49 sin30 = 49 * 1/2 = 24.5

Now, at any time later

v = v0 - at

s = 0 + v0*t - 1/2 a*t^2

= 24.5t - 4.9t^2

= 4.9t(5-t)

So, the arrow is at height=0 at time t=0 and t=5. Therefore, it is at max height when t=2.5

s(2.5) = 24.5*2.5 - 4.9*(2.5)^2 = 30.625 m

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