Kinematics Question

[sxld_cu]

This question really does not seem to make any sense to me
It is from the Kaplan Quiz Bank
Does anybody else have any perspectives on this?

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[Cawolf]

That's a really poor question IMO.

I think it is just best to picture the projectile flight path, thrown from some height.

The kinematics equations show that the half way point of x travel (horizontal), the projectile will be at the half way point of z travel (vertical).

From this you could deduce that for the projectile to be at 2 meters in height (half the initial) at 3 meters, it will need to go twice as far - 6 meters.

 

[type12]

agreed with Cawolf: not MCAT-esque, poorly written, and open to interpretation, which MCAT questions are not.

 

[justadream]

@Cawolf


"The kinematics equations show that the half way point of x travel (horizontal), the projectile will be at the half way point of z travel (vertical)."

Can you show this explicitly?

 

[Cawolf]

I went back and tried to relate what I said to the problem again - and it may not have been correct.

The problem requires some assumptions about the path of travel and initial velocity components. I still would use the initial reasoning I posted to come to an answer - but I don't think that the statement I made about the equations is technically correct. Apologies.

 

[Chrisz]

This question actually assumes that the monkey throws the ball horizontally. The vertical fall only depends on gravity. 1 meter fall would require 0.5(10)t^2=1, so t=sqr(0.2) horizontal speed would be 3/sqr(0.2). The entier fall would rquire t=sqr(4/5)=sqr(0.8). The horizontal distance would be x=vt={3/sqr(0.2)}sqr(0.8)=6m

 

[Chrisz]

Ignore the above answer, I made a mistake solving the question. The differencee between the head and the tree is actually 2 meters. I should also not assumed the monkey throw the ball horizontally. If these assumptions are made and the correct difference used, t=sqr(2/5)=sqr(0.4). Horizontal speed would be 3/sqr(0.4), and total distance would be {3/sqr(0.4)}sqr(0.8)=3sqr(2)=4.2

 

[Cawolf]

I followed that same path and you can see it doesn't align with the given answer choice of a x displacement of 6 m.

 

[Chrisz]

@Cawolf,
I think the reasoning should be the monkey throw the ball directly aiming at the monkey's head with a huge initial velocity that the gravity is ignorable, so there is going to be not a significant change in velocity(ignorable). This would make almost a linear trajectory. In this way, we can connect the trajectory both the monkey's head at 3m and ground at 6m

 

[Cawolf]

That makes sense being linear.

I kept picturing a parabolic path.

 

[The Brown Knight]

First, I think the question is approximating the path of the coconut as linear (it'd have to be thrown pretty fast for that!). Assuming that, it's just geometry:
Draw a trapezoid with the monkey, the base of the tree, and the 2 m tall person. Complete the triangle by joining together the lines from the monkey to the human's head and from the base of the tree to the human's feet to SOME point behind the human. (I swear this is much easier if you draw it out)
You'll see a "triangle within a triangle" relationship; because the triangles are similar, the following relationship holds:
Height of Human/Height of Tree = Length from tree to human/Length from tree to the point behind human
2/4 = 3/x
x = 6 m
Sorry for any confusion, but drawing it out will be much clearer!