How Fast Must an Object Travel to Reach 80 Meters High?

AI Thread Summary
To determine the minimum positive initial velocity required for an object to reach a height of 80 meters, the kinematic equation vf² = v₀² - 2gy is used, where vf is the final velocity (0 at peak height), g is the acceleration due to gravity (9.8 m/s²), and y is the height (80 m). By rearranging the equation, the initial velocity can be calculated as v₀ = √(2gy), leading to v₀ = √(1568), which approximates to 39.6 m/s. Among the multiple-choice options, 40 m/s is the closest and correct answer. The discussion confirms that the calculations are valid and that the assumptions made are reasonable for this scenario.
ch3570r
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This seems a simple enough question, but I am not sure on how to go about solving it

"What is the minimum positive initial velocity necessary for an object to be launched from ground level so that it reaches a height of 80 meters?"

Multiple choice answers: a) 5m/s
b) 9.8m/s
c) 12.5m/s
d) 40m/s
e) 80m/s

Well the problem doesn't give me a time, so I am guessing it is asking that the object reaches at least 80m before it starts its fall back down. Can someone attempt to explain how I would get the answer? If I were simply guessing, I would go with either C or D, however, that's only guessing. I am sure gravity would someone used, but I don't know what to do with it.
 
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Let's see, we are given explicitly the maximum height of the object as 80 meters. If we make the assumption the object is close to a flat Earth approximation, then the acceleration acting on the object is -g. Implicitly, we can that if the object just goes the minimum height of 80 meters, then its final velocity would be zero. Kinematically, this gives us the relation:

vf2 = v02 - 2gy

This is assuming it is thrown straight up. Would the initial velocity be greater or less if the object is thrown up at an angle?
 
Ok, so using the formula vf2 = v02 - 2gy, I can plug in the numbers as such: 0 = x^2 - 2(9.8)(80) where x is the initial velocity. This reduces to x^2 - 1568. Now, I am assuming that I can simply plug in the choices for x, which means I can square each choice, and if it is greater than 1568, it will reach 80m.

Choice D, 40m squared will give me 1600, which is the closest to 1568 and the minimum initial velocity.

Does that sound right, or did I just stack assumptions to get my answer?
 
Looks correct to me.
 
D is correct, but I think you are making it more difficult that it actually is. The given kinematic equation only has one unknown, v02. Just isolate the variable and take the square root of both sides.
 
you're right physics.guru, when I isolated the variable, it came out to 39.6, which is close enough to 40.
 
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