height/time question: Real gmat question?

[zzhop]

An object thrown directly upward is at a height of y feet after t seconds, where y = -16 (t - 3)^2 + 150. At what height, in feet, is the object 1 second after it reaches its maximum height?

 

A. 6

B. 86

C. 134

D. 150

E. 214

[bombay_bronco]

Take the derivative of y with respect to t to find WHEN the object reaches max ht.

 

after that plug in t+1 for t to find where it reaches in ht.

 

So in the above eqn, y is max when dy/dt=0

ie

dy/dt = -16(2)(t-3) + 0 = 0 ==> t = 3

 

no substitute t = 4 in above eqn to get -16(1) + 150 = 134.

Whats the OA?

 

So answer should be C - 135

 

Sorry C: 134

(my fingers are too stubby for this mac keyboard)

[zzhop]

Let others take a crack at it before I give the OA out. I want to see more explanations

[nenuphani]

Max height is reached when t=3.

i.e. max height is 150.

 

so, when t=4, what is the height? This is the ques asked indirectly.

 

Ans is 134

[Ankost]

Agree.

y = -16 (t - 3)^2 + 150 is complete square equation, where 3 is the max t and 150 - max height or vertex of quadratic function.

After 1 sec 3+1=4 sec substitute 4 in the equation and get 134.

[yoda_ngen]

ACtually, I also used the derivitive to find the time for max height. which came out to be t =3 and then, used t =4 in the given formula to get a height of 134 feet.

[RotiDaal]

I agree with all the solutions that have been given here. But I would like to add another way of viewing the equation, which might be helpful for some other people.

 

There are two terms in the equation: "-16 (t - 3)^2" and "+150" ... just observe that the first term will always be negative, no matter what the value of "t" ... so to max the value of the expression we have to minimise the value of this negative term that will be reduced from "+150" ... and this min value is 0 ... got by making t=3 ... so ... there you go ... two essential values are obtained in one step .... i.e ... max height is 150 and it is attained at t=3. Now all that one has to do is add the required time duration to t=3 and substitute it back into the eq which would be t=3+1 in this case.

 

Slightly off the topic:

Personally, I think the problems given on the GMAT will rarely expect us to differentiate/integrate any expression. And I think that the expressions are also limited to quadratic ones at max. So its good to remember the curve that one would obtain when a quadratic eq is plotted. The curve is some form of a symmetric parabola. Which means that the value of "y" reduces, reaches a min and then increases or vice-versa ... i.e ... value of "y" increases, reaches a max and then reduces.

 

Trying to see if this concept has any relevance when you encounter problem with a quadratic eq can help in a lot of situations.

[ChhotiAankhe_BadeSapane]

I don't think we need to apply derivative and calculus here. That is not in the GMAT Scope if this is GMAT Q.

 

From the equation you can see that, the maximum value of y is when t = 3.

 

We can say that after 3 secs, the obj reaches max height.

 

That's it, now put t = 4 and you get 134. That's the answer.

[venom]

The nature of the equation makes it very simple to identify the time taken to reach maximum height. Its at t=3.

 

Answer should be 134.