SOLUTION: Sam shot his model rocket up into the air and counted the seconds it stayed in flight. He noticed the rocket seemed to be at the same height as the top of a television antenna at 3
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: Sam shot his model rocket up into the air and counted the seconds it stayed in flight. He noticed the rocket seemed to be at the same height as the top of a television antenna at 3
Log On
Question 188501: Sam shot his model rocket up into the air and counted the seconds it stayed in flight. He noticed the rocket seemed to be at the same height as the top of a television antenna at 3 and 6 seconds.(gravity of earth is 32ft/s)
a)find initial speed of rocket
b)height of television antenna. Answer by solver91311(24713) (Show Source):
The first thing is that the acceleration due to gravity is . The minus sign is because Sam is shooting his rocket upward and the force of gravity acts in the opposite direction.
Now we need a function that represents the height as a function of time. Acceleration is the rate of change of velocity, so the instantaneous velocity as a function of time is given by:
where the constant is the initial velocity, so:
Velocity is the rate of change of distance (read height for this problem) so:
where the constant is the initial height, so:
However, one must presume (because you didn't say but it seems reasonable) that Sam launched his rocket from the ground or very near to it, so we can assume that the initial height, .
Giving us:
Into which we substitute the value of acceleration due to gravity:
Now, we know that the height was the same at 3 seconds and at 6 seconds, so we can say:
Solving for
Which answers part a.
Part b.
Now that we know the initial velocity, we can evaluate the height function at either 3 or 6 seconds to determine the height of the antenna.
