Finding the initial velocity of a thrown object caught at a certain height

AI Thread Summary
To determine the stopping distance of a car traveling at 35.1 m/s with an acceleration of -2 m/s², the equation vf² = Vi² + 2ad can be used, where vf is the final velocity (0 m/s), Vi is the initial velocity (35.1 m/s), and a is the acceleration (-2 m/s²). The calculated stopping distance is approximately 308 meters. Additionally, for the rocket problem, when thrown straight up at 12.42 m/s and caught 5 meters above the launch point, the final velocity can be determined using kinematic equations. The user expresses a need for clarification on the symbols and equations involved, indicating a desire to refresh their physics knowledge for an upcoming event. Overall, the discussion revolves around applying kinematic equations to solve motion-related problems.
Jake M
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A car is traveling along a road at 35.1 meters/second(m/s). All of a sudden the navigator yells "Stop! The driver quickly applies the brakes at an acceleration of -2m/s/s (meters per second squared) and comes to a stop.

How much distance it take to stop the vehicle?

My question is what equations to use in solving this problem.
Thanks for any hep in advance.
 
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Givens:
Vi=35.1 m/s
Vf=0 m/s
a= -2 m/s/s
d= ?

THis is the formula you can use to solve for distance (d):
vf^2 = Vi^2 + 2ad
 
Hi there PM,
thanks for the info. It's been about 20years since I've had a physics problem to solve. This isn't home work or anything but rather practice for an up coming Rocketry event. Can you show me how to solve for this? The symbols have me abit confused. Also another word problem that came up was the following:
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.
 
Sorry PM, For the first equation my answer was 308.00250000000005
Am I on the right track?
many thanks again,
Jake.
 
yes that looks correct to me.
 
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?
 

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