Posted by Anonymos on .
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?

Physics 
Steve,
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 
Physics 
Anonymos,
a) h(t)=5t^2+15t+50
h(0)=0+0+50
h=50