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September 1, 2014

September 1, 2014

Posted by **Julia** on Thursday, December 8, 2011 at 9:24pm.

Find the time when the projectile has returned to the initial height launch, find the time the projectile hits the ground, find the time when the projectile hits the ground when its velocity is 0, find the height of the projectile above ground after 1 second

- Algebra II -
**Damon**, Thursday, December 8, 2011 at 9:42pmg = -9.8 m/s^2

v = +20 - 9.8 t

y = 10 + 20 t -4.9 t^2

a) when is y = 10 again?

10 = 10 + 20 t -4.9 t^2

t(4.9 t-20) = 0

t = 20/4.9 = 4.08 seconds

b) when is y = 0 ?

4.9 t^2 - 20 t - 10 = 0

t = [20 +/- sqrt (400+196)] / 9.8

forget negative time, that was before we started, use + sign

t = 4.53 seconds

c) velocity is 0 at the op when

0 = 20 -9.8 t

t = 2.04 seconds

d) y = 10 + 20 (1) -4.9 (1)^2

= 25.1 meters

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