Proportionality

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bml1105: Master | Next Rank: 500 Posts; Posts: 113; Joined: Sun Nov 24, 2013 9:19 pmFollowed by: 5 members; Thanked: 1 times

by bml1105 » Mon Dec 23, 2013 7:56 pm

The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a max height of 9 feet?

How can you tell which term (16 or 4) is x or y?

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Mon Dec 23, 2013 9:36 pm

h = k * v^2

v = 16, h = 4

4 = k * 16^2

k = 4/256

h = 9, v = ?

9 = (4/256) * v^2

9 * 64 = v^2

v = 3 * 8 = 24

R A H U L

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Mon Dec 23, 2013 10:42 pm

bml1105 wrote:The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a max height of 9 feet?

How can you tell which term (16 or 4) is x or y?

Read the question carefully, then decide if your x is velocity or height, y would be height or velocity accordingly.

When two quantities bear direct proportion, their ratio is constant. If v is the velocity with which the object is thrown and h is the maximum height reached by the object thrown directly upward, then we may have

h1/(v1)^2 = h2/(v2)^2.

In the presented question, we numerically have

h1 = 4, v1 = 16, and h2 = 9.

From the relation, we can have

4/(16)^2 = 9/(v2)^2

or (v2)^2 = 9 (16)^2/4, and hence v2 = [spoiler]3 (16)/2 = 24 feet/sec[/spoiler].

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Dec 24, 2013 6:08 am

bml1105 wrote:The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a max height of 9 feet?

Let h = the maximum height and v = the velocity.

The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown.
Implication:
hâ‚�/(vâ‚�)Â² = hâ‚‚/(vâ‚‚)Â².

If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a max height of 9 feet?
4/16Â² = 9/vÂ²
4vÂ² = 16Â² * 9
vÂ² = 16 * 4 * 9
v = 4 * 2 * 3
v = 24.

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