Math
posted by NeedHelp .
The height(h) of an object that has been dropped or thrown in the air is given by: h(t)=4.9t^2+vt+h
t=time in seconds(s)
v=initial velocity in meters per second (m/s)
h=initial height in meters(m)
A ball is thrown vertically upwardd from the top of the Leaning Tower of Pisa (height=53m) with an initial velocity of 30m/s. Find the time(s) at which:
a) the ball's height equals the hight of the tower
b) the ball's height is greater than the height of the tower
c) the ball's height is less than the height of the tower
d)the ball reaches its maximum height
I don't know how to do this problem.
Please Help and Thank You very much =)

heck, they gave you the equation.
h(t) = 4.9t^2 + 30t + 53
a) ball goes up, comes back down to the top of the tower. So, we want
53 = 4.9t^2 + 30t + 53
0 = 4.9t^2 + 30t
0 = t(4.9t + 30)
so, t=0 (at the start) or t = 6.12 (as it comes back down)
b) same calculation but, t is between 0 and 6.12. That is 0 < t < 6.12
c) same calculation, but restricting t to positive values, t>6.12
Naturally, we could also restrict t to the point where height >= 0.
d) vertex of any parabola is where x = b/2a = 30/9.8 = 3.06
h(3.06) = 98.9 
I don't get how you got:
vertex of any parabola is where x = b/2a = 30/9.8 = 3.06
h(3.06) = 98.9
where did all the numbers come from? 
how did you get t=6.12 for a)???
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