1. ## Pre-calc question?

Hey!

A model rocket is launched. The height, in feet, of the rocket h(t) at t (t>=0) seconds after launch is determined by the equation h(t)= (-1/2t)^2 + 15t. The maximum height of the rocket is 112.5 feet.

(a) Find the number of seconds after launch it takes for the rocket to reach its maximum height. [Show all work]

(b) For how many seconds will the rocket be at a height of more than 100 ft? [Show all work]

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ok so its not like i tried it, i did, and let me show what i did.

first i plugged 112.5 for h(t) and got an answer of t=15 for part a.

for part b i plugged in 100 for h(t) and got 10 and 20. but i rejected 20 because thats higher than 15 which is the amount of time to reach the maximum height so that makes no sense. so i did 15 - 10 and got 5 for part b.

am i correct?

2. ## Re: Pre-calc question?

Hello, Hell0!

A model rocket is launched.
The height, in feet, of the rocket h(t) at t (t>=0) seconds after launch
is determined by the equation: . $h(t) \:=\: \text{-}\tfrac{1}{2}t^2 + 15t.$
The maximum height of the rocket is 112.5 feet.

(a) Find the number of seconds after launch it takes for the rocket to reach its maximum height. [Show all work]

(b) For how many seconds will the rocket be at a height of more than 100 ft? [Show all work]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let me show what i did.

First i plugged 112.5 for h(t) and got an answer of t = 15 for part (a). . Correct!

For part (b) i plugged in 100 for h(t) and got 10 and 20. . Right!
But i rejected 20 because thats higher than 15
which is the amount of time to reach the maximum height. . This is incorrect.

The rocket is on its way UP aand reaches 100 feet when t = 10 seconds.

It reaches maximum height when t = 15 seconds.

Then it is on its way DOWN.
It is 100 feet above the ground (again) when t = 20 seconds.

The rocket is above 100 feet from t = 10 to t = 20.

3. ## Re: Pre-calc question?

i dont really understand the going down part

4. ## Re: Pre-calc question?

Originally Posted by Hell0
Hey!

A model rocket is launched. The height, in feet, of the rocket h(t) at t (t>=0) seconds after launch is determined by the equation h(t)= (-1/2t)^2 + 15t. The maximum height of the rocket is 112.5 feet.

(a) Find the number of seconds after launch it takes for the rocket to reach its maximum height. [Show all work]

(b) For how many seconds will the rocket be at a height of more than 100 ft? [Show all work]

----------------------

ok so its not like i tried it, i did, and let me show what i did.

first i plugged 112.5 for h(t) and got an answer of t=15 for part a.

for part b i plugged in 100 for h(t) and got 10 and 20. but i rejected 20 because thats higher than 15 which is the amount of time to reach the maximum height so that makes no sense. so i did 15 - 10 and got 5 for part b.

am i correct?
I would sketch a graph of h(t). Use a graphing calculator if you need help.