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A toy rocket is launched vertically from ground level \((y = 0 \, \text{m})\) at time \(t = 0.0 \, \text{s}\). The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 64 m and acquired a velocity of 60 m/s.

The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground.

The time interval (in seconds) during which the rocket engine provides upward acceleration is closest to:

Answer :

Answer:

2.13 s

Explanation:

Hi!

At t = 0s the rocket is at rest in its platform, so the intial speed is zero. I f the acceleration is A, then the height Y, and the speed V are:

[tex]Y=\frac{A}{2}t^2[/tex]

[tex]V=At[/tex]

We nedd to find time T during which the rocket engine provides upward acceleration. We know that:

[tex]64\;m=\frac{A}{2}T^2\\ 60\frac{m}{s} =AT\\[/tex]

With these 2 equations we can find A and T (dropping units for simplicity):

[tex]A=\frac{60}{T} \\64 =\frac{30}{T} T^2=30T\\T=\frac{64}{30}\approx 2.13\;s[/tex]

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