Physics question, Tricky or wrong answer?

[cleopetra]

Hi !!!
can any one help me on the following question:
If an antilope is running at a speed of 10 m/s, and can maintain that horizontal velocity when it jumps, how high must it jump in order to clear a horizontal distanceof 20 m.
A. 5m
b. 10m
C. 20m
D. 45m

Answer is (for answer highlight) 45m. Can somene plz explain it?? I thought it would be 5m.

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[Kaustikos]

Hi !!!
can any one help me on the following question:
If an antilope is running at a speed of 10 m/s, and can maintain that horizontal velocity when it jumps, how high must it jump in order to clear a horizontal distanceof 20 m.
A. 5m
b. 10m
C. 20m
D. 45m

Answer is (for answer highlight) 45m. Can somene plz explain it?? I thought it would be 5m.

Where did you get that question from? Is there an explanation?
I assumed 10 m/s for 20 m = 2 seconds. Are we not assuming Vf = 0?
I'm guessing it is a trick question. How the hell would you have the same initial Vx without some outside source? Some of it has to transfer in the jump.

 

[RSAgator]

I got 5 as well. Unless units are wrong or something the answer is prolly wrong.

If he needs to travel 20m and he is traveling at 10m/s he needs to be in the air for 2s. Since he spends the same time going up as he does coming down it's 1s up 1s down.

x = (1/2)*a*(t^2 )

a = 10, t = 1

x = 5

 

[BerkReviewTeach]

Hi !!!
can any one help me on the following question:
If an antilope is running at a speed of 10 m/s, and can maintain that horizontal velocity when it jumps, how high must it jump in order to clear a horizontal distanceof 20 m.
A. 5m
b. 10m
C. 20m
D. 45m

Definitely a typo. The math listed in the explanations above is correct.

As an alternative solution, I'll propose a short cut that I present in lecture. Draw two 45-45-90 triangles flush in such a way where the overall triangle is 45-90-45 triangle that is a in height and 2a in the base. If you draw a parabolic curve from the lower left corner to the lower right corner, you'll find that it approximately bisects the vertical line. This means that when the object is launched at 45 degrees, it's range is four times its apex. That should apply to this question, and any other range question where the launch angle is 45 degrees and the landing point is at the same elevation as the launch point.

 

[DaTruMD]

Definitely a typo. The math listed in the explanations above is correct.

As an alternative solution, I'll propose a short cut that I present in lecture. Draw two 45-45-90 triangles flush in such a way where the overall triangle is 45-90-45 triangle that is a in height and 2a in the base. If you draw a parabolic curve from the lower left corner to the lower right corner, you'll find that it approximately bisects the vertical line. This means that when the object is launched at 45 degrees, it's range is four times its apex. That should apply to this question, and any other range question where the launch angle is 45 degrees and the landing point is at the same elevation as the launch point.

That was interesting. Do you have any more "short cuts"?

 

[RickAstley]

OK now wtf antelope is going to leap 20 METERS? 10 m/s is reasonable, but 20 meters is like 65 feet, amirite?

 

[cleopetra]

Where did you get that question from? Is there an explanation?
I assumed 10 m/s for 20 m = 2 seconds. Are we not assuming Vf = 0?
I'm guessing it is a trick question. How the hell would you have the same initial Vx without some outside source? Some of it has to transfer in the jump.

This Question was on EK physics review ,, Chapter1- question 19

 

[Kaustikos]

OK now wtf antelope is going to leap 20 METERS? 10 m/s is reasonable, but 20 meters is like 65 feet, amirite?

Apparantly we can take away logic and sense for mcat questions.

 

[Vihsadas]

Definitely a typo. The math listed in the explanations above is correct.

As an alternative solution, I'll propose a short cut that I present in lecture. Draw two 45-45-90 triangles flush in such a way where the overall triangle is 45-90-45 triangle that is a in height and 2a in the base. If you draw a parabolic curve from the lower left corner to the lower right corner, you'll find that it approximately bisects the vertical line. This means that when the object is launched at 45 degrees, it's range is four times its apex. That should apply to this question, and any other range question where the launch angle is 45 degrees and the landing point is at the same elevation as the launch point.

Hey cool trick. I never thought of that one before...

 

[BerkReviewTeach]

Hey cool trick. I never thought of that one before...

There are a bunch of simple little geometry and trigonometry tricks that really help with physics, especially in optics and translational motion. Too bad they ask so few questions on the MCATs, because about 90% of the tricks I show students end up not being covered on their exam.