EK Physics Questions 102

[Violagirl]

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I'm having a hard time with this free fall question from EK physics:

An object is dropped from a height h and strikes the ground with a velocity v. If the object is dropped from a height of 2h, which of the following represents its velocity when it strikes the ground?

A) v

B) 1.4v

C) 2v

D) 4v

The solution says to use the equation, v = sqrt (2gh). I'm really confused on where this equation comes from for this problem and am not quite sure how to apply it. Any help is appreciated, thanks so much!

 

[premed1001]

this is part of the 5 translational motion equations that you must memorize

Vfinal^2=Vinitial^2 + A2D

plug in Height into D vs plug in 2*height into D and then solve for V. You will see.

V initial is 0 since it is being dropped

A is 9.8 AKA 10

 

[WolfofSDN]

No, there

this is part of the 5 translational motion equations that you must memorize

Vfinal^2=Vinitial^2 + A2D

plug in Height into D vs plug in 2*height into D and then solve for V. You will see.

V initial is 0 since it is being dropped

A is 9.8 AKA 10

No no no no no. This is way too long and complicated. Conservation of energy and a little intuition about proportions will take care of this problem. Who has time on the MCAT to write down one of the big 5 equations, plug everything in, then solve? hell no.

Here it is:
mgh = 1/2mv^2
Proportion:
h = v^2
so, when you double the height, you quadruple the velocity.

The answer is D, 4v

 

[dudewheresmymd]

No, there

No no no no no. This is way too long and complicated. Conservation of energy and a little intuition about proportions will take care of this problem. Who has time on the MCAT to write down one of the big 5 equations, plug everything in, then solve? hell no.

Here it is:
mgh = 1/2mv^2
Proportion:
h = v^2
so, when you double the height, you quadruple the velocity.

The answer is D, 4v

You made a math error.

OP, here is the solution to how they derive the sqrt gh:

Use energy conservation:

delta ke+deltape=delta ke+delta pe (delta KE is 0 on both sides even after doubling the height--> Key word is "dropped" meaning that the initial kinetic energy is 0), therefore set the potential energy of the object at the top = the kinetic energy of the object at the bottom of its fall (assuming no non conservative forces are acting, which is a valid assumption for this problem as they do not state otherwise)

mgh=1/2mv^2 (cross off m from both sidese)
gh=1/2v^2 (multiply both sides by 2)
2gh = v^2 (take square root of both sides)
sqrt (2gh) = v

Therefore, v (velocity) is directly proportional to sqroot of height.

If you double the height, the velocity increases by a factor of sqroot 2 or 1.41v. The correct answer is B.