Math problem!! Really confused!?

Question

Olive Math is standing on the 102nd floor of the Empire State building, thinking about mathamatics. She realizes that, if she were able to throw a ball from her present position, the height of the ball in feet, h, at time t seconds could be modeled by the equation
h = -16t^2 + 64t + 1224.

How high above the ground is the 102nd floor?
What is the greatest height that ball would reach?
How long would it take to reach that height?
How long would it take the ball to hit the ground?

Please help?? PLease just help me get started, im just really confused.

Answer 1

PROBLEM 1:
How high above the ground is the 102nd floor?

We assume that the ball starts at her location on the 102nd floor. So at time t = 0, that will give the height of the 102nd floor.

I believe you get h=1224, if you plug in t = 0

PROBLEM 2:
What is the greatest height that ball would reach?

If you know calculus, take the derivative and solve for zero. If you don't solve the equation in vertex form.

Start by pulling out -16 so you have t² by itself:
h = -16(t² - 4t - 76.5)

Now complete the square by taking the coefficient on the t term (-4), halve it (-2), and square it (4). Add this and subtract it inside the parentheses:

h = -16(t² - 4t + 4 - 4 - 76.5)
h = -16(t² - 4t + 4 - 80.5)

Get the -80.5 outside the parentheses, by multiplying by -16:
h = -16(t² - 4t + 4) -16(-80.5)
h = -16(t² - 4t + 4) + 1288

Write the parentheses now as a perfect square:
h = -16(t - 2)² + 1288

The greatest height will be when we subtract nothing. In other words if the squared iterm were equal to 0, then we would subtract nothing. The maximum height would then be 1288 feet.

PROBLEM 3:
How long would it take to reach that height?

h = -16(t - 2)² + 1288

Again, when the squared term is zero, then we subtract nothing. That would be the maximum height.

Solve:
(t-2)² = 0

Take the square root of both sides:
t - 2 = 0
t = 2

So the ball will be at the maximum height 2 seconds after being thrown. Then it will start going back down.

PROBLEM 4:
How long would it take the ball to hit the ground?

Solve for t and figure out when h = 0. The quadratic formula will work for this... or you can solve the equation below for 0:

h = -16(t - 2)² + 1288
0 = -16(t - 2)² + 1288

Add 16(t - 2)² to both sides:

16(t - 2)² = 1288

Divide both sides by 16:
(t - 2)² = 1288/16
(t - 2)² = 80.5

Take the square root of both sides:
t - 2 = √(80.5)
Add 2 to both sides:
t = 2 + √(80.5)
t ≈ 10.9721792 seconds.
Rounding to the nearest hundredth:
t ≈ 10.97 seconds

Answer 2

first you should calulate how long it take to reach the ground. that is h = 0.
that will give you a positive and an negative t in seconds.

next you should inverse the direction by throwing the ball in the air with the same max speed that you get when the ball falls.
this max speed occurs when -2*16 t + 64 = 0 or at t = 2 seconds. ( which is also the time it took to reach the floor )

so fill in the 2 seconds and voila the h follows h = -16 * 2*2 + 64*2 + 1224 = 1288

( i am assuming the ball is thrown vertically )

update
its a bit more complicated ......
when the ball is thrown the h is first increasing thus the ball is not thrown vertically

any way it will take 2 seconds to reach the max height. which is 1288 feet then.

next the ball falls until h = 0
this take 0 = -16t*t + 64t + 1224 => gives some t

well you need to subtract the height and the times and notice that g = 9.8 usually

good luck

h = 0

Answer 3

1) at time zero Olive is standing on the 102nd floor of the building so from formula h(t=0)=1224 feet

2,3) if you know differentiation its easy to say that. you should just make a differentiation from the formula and equal it to zero as follows: h'=-32t+64=0 so t = 2 sec
and the height is 1288 feet

if you don't know differentiations. you might know the formula for the Max. value of the curve y=ax^2+bx+c which is :
x=-b/2a y= (-b^2-4ac)/4a

from these the height will be 1288 feet and time will be 2 sec.

4) for calculating this, when ball hits the ground the h will be zero so by calculating the equation for h=0 we will find t
t=-6,97sec or t=10,97sec as you there is no negative time so the the answer is t=10,97sec

Answer 4

When t =0 h =1224 so that's the height of 102nd floor

-b/2a = -64/(2*-16) = 2 seconds, so max height is;
-16(2^2) +64(2) +1224 = 1352 feet = max height.
It would take 2 seconds to raech that height

-16t^2 +64t +1224 = 0
t = [-64 +/- sqrt(64^2 - 4(-16)(1224))]/(2*-16)
t = approximately 10 seconds to hit ground.

Answer 5

S = So + Vot + [a*t^(2)] / 2

1) Then, So (the initial space or the height) = 1224 feet.

4) Vo = 64, a = -16.

V = Vo + a*t

0 = 64 -16*t

t = 64/16 = 4 seconds.

Answer 6

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Answer 7

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Answer 8

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