 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
A model rocket is launched at an upward velocity of 460feet per second off a platform that is 40 feet up from the ground. Use the formula h(t)= -16+vo+ho where vo is the intial velocity an ho is the intial hight. A.) Find the maximum hight of the rocket. B.) How long does it take to reach the maximu height? C.) When will the rocket hit the ground? |
Hi Megan.
I think you meant to write the formula as
h(t) = -16t2 + v0 t + h0
We ignore air resistance for this question, so we know that if the rocket takes 20 seconds to go all the way up and come back down to the level of the launch, then at time t=10 seconds it was at the maximum height. So if we can determine how long it takes to return to the level of the platform, we can answer the question (B). So h(t) = h0 when the rocket returns. That means
h0 = -16t2 + v0 t + h0 so
0 = -16t2 + v0 t and if we divide by t on both sides then
0 = -16t + v0 which simplifies to
t = v0/16
and since v0 = 460, then t = 460/16 = 28.75 seconds. At half that time, the rocket has reached the apex of its flight. This is the answer to part (B).
Now you want to calculate h(t) when t = (your answer to part B), v0 = 460 and h0 = 40:
h(t) = -16t2 + v0 t + h0
That will answer part (A) of your question.
For part (C), use the same equation. You want to find t when you know that h(t) = 0 (that's ground level), v0 = 460 and h0 = 40. You may need to use the quadratic formula to find the time t.
Hope this helps,
Stephen La Rocque.
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.