Tricky word problem

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moronatmath

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A ball is thrown upward from the top of a building 200 feet tall.
The height of the ball is described by the function h(t) = - 16 t^2 +18 t + 200, where t is in seconds and t = 0 corresponds to the moment the ball is thrown.
a) Determine for what value of t the ball reaches the maximum height and determine this maximum height.

t =

maximum height =
b) Determine when the ball reaches the ground.

t =


Ok the only thing I know how to do in this problem is find the max height. I think it is 205.0625 .

Can I get some help with this problem please?
 
You have the max. height correct. When it hits the ground the elevation is 0. Find the zeros of the given quadratic. There will be one positive and one negative. Since you can't go back in time, I would use the positive one.
 
Wehn I plug in T i get 200.

The solution 0 for when the ball hits the ground was incorrect. :(

How do I find the solutions to 0?
 
moronatmath said:
Wehn I plug in T i get 200.
Click to expand...
There is no "T" in the exercise, as posted. Or did you mean "t"...? And you're supposed to be finding the value of t. What do you mean when you say you plugged in the value for t?

moronatmath said:
The solution 0 for when the ball hits the ground was incorrect.
Click to expand...
I'm sorry, but I don't understand what you mean by this. The value of the height h is indeed zero when the ball is on the ground (for part (b) of the exercise).

Please reply with clarification. Thank you.

Eliz.
 
stapel said:
moronatmath said:
Wehn I plug in T i get 200.
Click to expand...
There is no "T" in the exercise, as posted. Or did you mean "t"...? And you're supposed to be finding the value of t. What do you mean when you say you plugged in the value for t?

I meant if I do t=0 I get 200. M


moronatmath said:
The solution 0 for when the ball hits the ground was incorrect.
Click to expand...
I'm sorry, but I don't understand what you mean by this. The value of the height h is indeed zero when the ball is on the ground (for part (b) of the exercise).

Please reply with clarification. Thank you.

Eliz.
Click to expand...

Well I entered 0 for part B of the problem and was incorrect. I guess I don't understand what I am supposed to do when finding the zeros of the given quadratic?
 
moronatmath said:
Well I entered 0 for part B of the problem and was incorrect.
Click to expand...
Plugging in the value for the height does not answer the question. You were asked to find the time at which the height is zero. You still need to solve the equation.

Eliz.
 
Just solve the quadratic.

\(\displaystyle {-16}t^{2}+18t+200=0\), solve for t using the quadratic formula, completing the square, whatever strikes your fancy.

Please don't say, "How do I do that"?. :wink:
 
The height of the ball is described by the function h(t) = - 16 t^2 +18 t + 200, where t is in seconds and t = 0 corresponds to the moment the ball is thrown.
a) Determine for what value of t the ball reaches the maximum height and determine this maximum height.
t=
maximum height =
b) Determine when the ball reaches the ground.
t =
Ok the only thing I know how to do in this problem is find the max height. I think it is 205.0625 .
Can I get some help with this problem please?

1--From the given equation for height, the initial upward velocity must be 18 ft/sec based on the equation for height of h = Vot - gt^2/2 where Vo = the initial velocity, g = the deceleration due to gravity and t = the time to reach the height h with velocity zero..
2--From Vf = Vo - gt, (Vf = the final upward velocity = 0) we get o = 18 - 32t making t = .5625 sec.
3--Then, h = 18t - 16t^2 which leads to h = 5.0625 ft.
4--Thus, the total height reached is 205.0625 ft. above the ground.
5--When the ball starts down, Vo = 0 giving us 205.0625 = 0(.5625) + 16t^2 from which t^2 = 12.816 or t (fall to ground) = 3.58 sec.
6--The toal time from the initial upward velocity of 18 ft/sec to the ground is therefore 4.142 sec.
 
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