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January 31, 2015

January 31, 2015

Posted by **Sam** on Wednesday, September 3, 2014 at 6:23pm.

A ball is thrown upwards, from a building. It's height(h), in meters above the ground seconds(t) after being thrown is given by h(t)=-5t^2+15t+50

a) How tall is the building? (explain your answer)

b) How long does it take the ball to reach the ground?

c) For how long is the ball 55m above the ground?

- 11 Physics -
**Steve**, Wednesday, September 3, 2014 at 7:58pmsurely you have similar question that you can study.

You know that the initial height is the constant term. So, the building is 50m tall

The ball hits the ground when the height is zero. So, just solve for t when h=0

I assume you want to know how long the ball is at least 55m above the ground. It is only at 50m for an instant. So, we want

-5t^2+15t+50 >= 55

1/2 (3-√5) <= t <= 1/2 (3+√5)

http://www.wolframalpha.com/input/?i=+-5t^2%2B15t%2B50+%3E%3D+55

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