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# Question that came up on the test

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Manager
Joined: 21 Mar 2006
Posts: 89

Kudos [?]: 2 [0], given: 0

Question that came up on the test [#permalink]

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28 Mar 2006, 09:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The following is a question that I confronted on the actual test the other day. Obviously, it is somewhat re-worded, but I am quite certain about the formula and the numbers:

The formula H=-16t^2+vt represents the height at which a ball is thrown in the air. V is the velocity in ft/second and T is the time. If a ball is thrown in the air a velocity of 64 ft/second, how long will it take for the ball to hit the ground?

Some of the available answer choices included: 8, 16, 32 and 0.

Kudos [?]: 2 [0], given: 0

Manager
Joined: 09 Feb 2006
Posts: 129

Kudos [?]: 8 [0], given: 0

Location: New York, NY

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28 Mar 2006, 09:23
This is a pretty standard physics problem.

The formula H=-16t^2+vt represents the height at which a ball is thrown in the air. V is the velocity in ft/second and T is the time. If a ball is thrown in the air a velocity of 64 ft/second, how long will it take for the ball to hit the ground?

First, we can plug in 64 for v, giving us: H=-16t^2+64t. Now when the ball hits the ground, we know that the height is 0, so we just have to solve the equation for 0.

0 = -16t^2+64t
16t^2 = 64t
t^2 = 4t
Now, normally we might not be able to do this, but since we know t is NOT 0, we can divide both sides by t.

therefore, t = 4

Kudos [?]: 8 [0], given: 0

Manager
Joined: 21 Mar 2006
Posts: 89

Kudos [?]: 2 [0], given: 0

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28 Mar 2006, 10:50
Thanks for the explanation, that's helpful

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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5034

Kudos [?]: 437 [0], given: 0

Location: Singapore

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29 Mar 2006, 07:21
H = -16t^2 + 64t

It hits the ground when h becomes 0.

t(64-16t) = 0
t = 0 (invalid) or t = 64/16 = 16/4 = 4

Kudos [?]: 437 [0], given: 0

VP
Joined: 29 Apr 2003
Posts: 1403

Kudos [?]: 30 [0], given: 0

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31 Mar 2006, 06:30
Yeah this is a simple Kinematics problem

typically, S=ut + 1/2 a(t^2) where S is the distance, u is the initial velocity, a is the acceleration and t is the time..

In this case S (or H) == 0.

Then you can simply work on the quaratic equation. the values of t should be 0 and some other root.

You will have 2 choose the non zero root.

Kudos [?]: 30 [0], given: 0

Intern
Joined: 03 Mar 2006
Posts: 19

Kudos [?]: 1 [0], given: 0

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31 Mar 2006, 20:50
Good one.. trick is you have to know the height will be zero when it hits ground.

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VP
Joined: 29 Apr 2003
Posts: 1403

Kudos [?]: 30 [0], given: 0

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31 Mar 2006, 22:20
ywilfred said
Quote:
t = 0 (invalid)

Is NOT invalid. When the object has not left the hand (just when it is launched) t==0.

Kudos [?]: 30 [0], given: 0

Manager
Joined: 21 Mar 2006
Posts: 89

Kudos [?]: 2 [0], given: 0

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31 Mar 2006, 23:07
Based on my recent test experience and what I read from others, I feel like they are asking more trick questions like this on the GMAT. Questions in which the calculation might be fairly simple however its a matter of making the connection that the H=0. Some of you might say, well yeah its obvious that H is 0. But in reality, when you are in a test environment and under time pressure, these sort of things might not always come to you as intuitively as you would think. Unfortunately, there is also little preparation one can do for these kind of problems, in some ways they are more like a critical reasoning problem in nature than like the traditional problem solving questions that you see in the official guide.

Kudos [?]: 2 [0], given: 0

VP
Joined: 29 Apr 2003
Posts: 1403

Kudos [?]: 30 [0], given: 0

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01 Apr 2006, 04:01
I totally agree! And the thing is none of the prep materials - Kaplan, PR, PP, OG have these kind of questions!

Kudos [?]: 30 [0], given: 0

01 Apr 2006, 04:01
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