The path of a bottle rocket being launched into the air and falling back to the ground can be modeled by the equation s=1/2 at^2+v_o t+s_o where s is the height of the object, a is the acceleration, vo is the initial velocity, and so is the initial height. For one launch, the following measurements were taken at t = 0.5 second the height was measured as 144 feet, at t = 1 second the height was 152 feet, and at t = 2.5 seconds the height was 128 feet. Use this information to answer the following questions.
What was the initial velocity of the rocket?
What was the initial height of the rocket?
What was the acceleration?
What was the maximum height the rocket reached?
When did the rocket hit the ground?
I am at a loss on how to get started.
To answer the questions, we can use the given measurements and the equation s = (1/2)at^2 + vo*t + so. Let's tackle each question step by step:
1) What was the initial velocity of the rocket?
To find the initial velocity (vo), we need to first identify an equation that allows us to solve for vo. Looking at the given equation, we can rearrange it to solve for vo:
s = (1/2)at^2 + vo*t + so
Rearranging the equation, we get:
vo*t = s - (1/2)at^2 - so
vo = (s - (1/2)at^2 - so)/t
Now, let's substitute the values from the measurements to solve for vo. At t = 0.5 second, the height (s) was measured as 144 feet. Plugging the values into the equation, we have:
vo = (144 - (1/2)a(0.5)^2 - so)/0.5
2) What was the initial height of the rocket?
To find the initial height (so), we can use the same equation. However, when t = 0, the term (1/2)at^2 becomes 0, simplifying the equation to:
s = vo*t + so
Rearranging the equation, we get:
so = s - vo*t
Using the measurements at t = 0.5 second, we have:
so = 144 - vo(0.5)
3) What was the acceleration?
To determine the acceleration (a), we need to compare the given measurements at different times. We can use the equation twice and set up a system of equations to solve for a. Using the first two measurements:
152 = (1/2)at^2 + vo*t + so
144 = (1/2)at^2 + vo*t + so
Notice that the terms with vo and so cancel out since they are the same in both equations. Subtracting the second equation from the first, we get:
8 = 0 + 0 + 0
This implies that the acceleration is not present in these measurements, so a = 0.
4) What was the maximum height the rocket reached?
The maximum height corresponds to the vertex of the rocket's trajectory. We can use the formula for the vertex of a parabola to find it. In this case, the vertex is given by t = -b/2a, where a = (1/2)a and b = vo.
Using the initial velocity found in question 1, we can substitute the value of vo into the formula:
t = -vo / (2 * (1/2)a)
5) When did the rocket hit the ground?
To determine the time when the rocket hit the ground, we need to find the value of t when the height (s) is equal to 0. We can plug this information into the equation and solve for t:
0 = (1/2)at^2 + vo*t + so
Now, you can substitute the values we found in the previous steps into the corresponding equations to get the final answers to each question.
you have three equations and three unknowns
... the unknowns are ... a, Vo, and So
144 = 1/2 a (.5^2) + Vo .5 + So
152 = 1/2 a (1^2) + Vo 1 + So
128 = 1/2 a (2.5^2) + Vo 2.5 + So
solve the system for a, Vo, and So
... use the resulting equation to find max height and flight time
... max height occurs at max time (on the axis of symmetry ... -b/2a)
... flight time is when s equals zero