Maximum height PS problem. Need expert help.

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aman88: Master | Next Rank: 500 Posts; Posts: 142; Joined: Sat Oct 20, 2012 3:31 amFollowed by: 3 members; Location: New Delhi; Thanked: 20 times

by aman88 » Wed Dec 12, 2012 12:32 pm

Q: An object thrown directly upward is at a height of h feet after t seconds, where
h = -16 (t-3)2+150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?

A. 6
B. 86
C. 134
D. 150
E. 214

OA B

I checked some previous posts on this question but couldn't understand much of the solution. Request experts to please help me out on this PS problem.

Thanks.

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Source: Beat The GMAT — Problem Solving | Wed Dec 12, 2012 12:32 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Dec 12, 2012 1:01 pm

aman88 wrote:Q: An object thrown directly upward is at a height of h feet after t seconds, where
h = -16 (t-3)^2+150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?

A. 6
B. 86
C. 134
D. 150
E. 214

The formula h = -16 (t - 3)^2 + 150 allows us to determine the height of the object at any time. For what value of t is -16(t-3)^2+150 maximized (in other words, the object is at its maximum height)?
It might be easier to answer this question if we rewrite the formula as h=150-16(t-3)^2
To maximize the value of h we need to minimize the value of 16(t-3)^2 and this means minimizing the value of (t-3)^2.
As you can see,(t-3)^2 is minimized when t=3.

We want to know the height 2 seconds AFTER the object's height is maximized, so we want to know that height at 5 seconds (3+2)

At t=5, the height = 150-16(5-3)^2
=150-16(2)^2
=150-64
=86
Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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puneetkhurana2000: Master | Next Rank: 500 Posts; Posts: 131; Joined: Wed Nov 14, 2012 2:01 pmFollowed by: 2 members; Thanked: 39 times

by puneetkhurana2000 » Wed Dec 12, 2012 2:20 pm

h = -16 (t-3)^2+150

Carefully observing we find the expression -16 (t-3)^2 will always be negative as (t-3)^2 is always positive.

So, h attains its maximum(150) at t=3, so eliminate D and E.

Now, after a total of 5 seconds (3+2) ...it will come down and new height = -16*(5-3)^2+150 = 86

Answer B

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The Iceman: Master | Next Rank: 500 Posts; Posts: 194; Joined: Mon Oct 15, 2012 7:14 pmFollowed by: 6 members; Location: India; Thanked: 47 times

by The Iceman » Wed Dec 12, 2012 10:17 pm

In calculating the instantaneous height, we are subtracting a +ve time dependent value from 150. Hence max height possible is 150. This occurs at t=3.

Substitute t=5, and find the value of h.

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