Story problem that I completed--is my answer correct?

[paradise737]

Hello, my first math assignment of the year was a story problem and I wanted to see if I've answered it correctly (I have my doubts):

Basically, a group of people want to know the following information: 1) how long it will take a firework to reach the top of its trajectory 2) what that trajectory is and 3) the horizontal distance the firework will travel. The group of people wants to explode the firework when it reaches its maximum trajectory.

The firework will initially be rising at 92 ft/second, and they'll be launching it off of a 160-ft tower. It will be launched at a 65 degree angle.

The book gives the following formulas (height=h, time=t, distance=d)

h(t) = 160 + 92t - 16t-squared

d(t) = 92t / tan65

If you plug in 3 seconds for t in the first equation, it gives the maximum trajectory, which is 292 feet. That takes care of the first two questions, assuming I'm correct.

It's the third question I'm worried about. Since the people will want to explode the rock at exactly 3 seconds, I thought I should plug in 3 seconds into the second equation. That gives a horizontal distance of 128.78 feet.

But if you plug in 3 seconds into the first equation, it gives the maximum trajectory. In actuality, I want to know how far the firework traveled when it hits the ground, not when its at maximum trajectory--right?

If you understood any of what I just wrote, can you help me out with this?

 

[galactus]

The firework will hit the ground when the height is 0.

Solve \(\displaystyle -16t^{2}+92t+160=0\) for t. That will be the time until it hits the ground.

Then, to find how far it traveled, horizontally, when it hit the ground, plug that time into the d(t) formula.

The time to max trajectory is 2.875 seconds at a height of 292.25 and a horizontal distance of 123.34, to be exact.

The maximum height can be found by differentiating or using \(\displaystyle x=\frac{-b}{2a}=\frac{-92}{2(-16)}=2.875\)

That is the formula for the x-coordinate (in this case, t) of the vertex of a parabola (the max trajectory). Plug that back into the original quadratic to find the height it traveled.