Last Physics Word Problem, Need Help

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A man throws a rocket straight up with an initial speed of 12.42 m/s, and it is caught at a height of 5.0 meters above the launch point. To solve for the rocket's final velocity when caught, the kinematic equations for projectile motion are utilized. The maximum height is calculated, and the time to reach that height is found to be approximately 0.636 seconds. Using this time, the final velocity when caught is determined to be about 6.23 m/s. Proper application of kinematic formulas is essential for solving such physics problems.
Jake M
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A man is standing at a launch site. The rocket doesn't liftoff. Frustrated s/he throws it straight up with a speed of 12.42 m/s. It is caught on the way down at a point 5.0 meters above where it was thrown, by a fellow rocketeer on the second floor, who doesn't want to see any harm come ot this wonderful rocket. The first man wants to figure out how fast the rocket was going when it was caught.
Again thanks for all your help, really rusty with this right now.

Any help on this one?
Need to start with the right equation & how to solve.
 
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Make a list of what you know. Initial velocity = 12.42 m/s. Final velocity = 0 (at the top). Acceleration = 9.81 m/s^2. Use your kinmatics formulas to find the maximum height. The chapter of your book that gave you that question should have the formulas listed.

Then figure out how far it dropped from its highest point (where velocity = 0) to the point 5 meters above the ground. (maximum height - 5). Use this new distance and another kinmatic formula to calculate the rocket's velocity at that point.
 


To solve this problem, we can use the equation for projectile motion, which is:

y = y0 + v0t + 1/2at^2

where y is the final height, y0 is the initial height, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (which we can assume to be -9.8 m/s^2).

In this case, we know that the rocket was thrown with an initial velocity of 12.42 m/s and it reached a maximum height of 5.0 meters. We also know that the final height (y) is the same as the initial height (y0) because the rocket was caught at the same height it was thrown from. Therefore, we can set up the equation as:

5.0 m = 0 m + 12.42 m/s * t + 1/2 * (-9.8 m/s^2) * t^2

Solving for t, we get t = 0.636 seconds.

Now, to find the final velocity (vf) when the rocket was caught, we can use the equation:

vf = v0 + at

Plugging in the values, we get:

vf = 12.42 m/s + (-9.8 m/s^2) * 0.636 s

Therefore, the final velocity of the rocket when it was caught was approximately 6.23 m/s.

I hope this helps! Remember to always start with the appropriate equation and plug in the given values to solve for the unknown variable. Good luck!
 
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