Math Help (URGENT X10)

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)

H=height
h=initial height
Is there a difference but, anyway I didn't make this clear on the last post.

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
H(t)=-4.9t^2+30t+53
H(t)=???

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.

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3. 17
1. acceleration downward due to gravity is 10m/s to be exact 9.8m/s
this is enough to get you to start thinking

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posted by Rami Dahabreh
2. 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)

If t(-4.9t+30)=0, either
t=0
or
-4.9t+30 = 0
That is, t = 6.12

If you can't solve a factored expression, you have some review to do.

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

you know from the quadratic formula that x = -b/2a ± sqrt(blah blah)
Parabolas are symmetric, so the vertex is midway between the roots, which are equally spaced around x = -b/2a

h(3.06) = 98.9

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posted by Steve

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