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September 18, 2014

September 18, 2014

Posted by **Herosugar** on Monday, October 8, 2012 at 10:42pm.

I have solved for all answers but this one.

How long would it take the object to reach a height of 14.1 ? For an object thrown upward there are two times it would be at any hieght. Give the going up answer, a comma, and give the going down answer.

How do we go about this?

- College Physics -
**Samuel**, Friday, October 12, 2012 at 5:56amAssuming that accelerations is constant,

Given: a=15 ms^-2, s=14.1 m

Applying kinematics equation,

v^2 = u^2 + 2as

Find for u since v is 0,

u= sqrt(2 x -15 x -14.1)

= 20.78

Solve for t,

s= ut + 0.5at^2

14.1= 20.78t + 0.5(15)t^2

//Solve quadratic eqn for t

t= 0.564(Ans), -3.33(not possible)

- College Physics -
**Anonymous**, Wednesday, December 18, 2013 at 5:41amSolving for t,

14.1=20.78t-.5(15)t^2

subtract using a -15m/s^2 (downward acceleration due to gravity)

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