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A toy rocket moving vertically upward passes by a 2.1 m-high window whose sill is 9.0 m above the ground. The rocket takes 0.15 s to travel the 2.1 m height of the window. What was the launch speed of the rocket? Assume the propellant is burned very quickly at blastoff. How high will the rocket go?
It’s all about energy.
The time-of-flight past the window gives you the average speed of the rocket (it falls out of a quadratic); since you also have the duration of the pass, you can calculate the speed at the bottom of the window.
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Once you have the speed at any point, you can calculate it at any other point by adding or subtracting mgΔh from the kinetic energy. To calculate the apogee, start with the kinetic energy at h=0 and set equal to mgh, then solve for h.
Here’s an example of your quadratic:
h(t) = v0t – ½gt²
h(t+0.15)-h(t) = 2.1, solve for v0
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It’s all about energy.
The time-of-flight past the window gives you the average speed of the rocket (it falls out of a quadratic); since you also have the duration of the pass, you can calculate the speed at the bottom of the window.
Once you have the speed at any point, you can calculate it at any other point by adding or subtracting mgΔh from the kinetic energy. To calculate the apogee, start with the kinetic energy at h=0 and set equal to mgh, then solve for h.
Here’s an example of your quadratic:
h(t) = v0t – ½gt²
h(t+0.15)-h(t) = 2.1, solve for v0
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