An object thrown directly upward

This isn't a GMAT problem, this is a calculus problem. We know that when an object reaches it's maximum height, it's velocity is zero. So to find the time the object is at it's maximum height, we take the derivative and set it equal to zero.

h'(t) = -32(t-3)

Set it equal to zero and solve for t and we find t = 3. 2 seconds after this time, t = 5.

We want to know the height at t=5. This is h(5) = -16(5-3)^2 + 150 = -16*4 + 150 = -64 + 150 = 86.

The answer is B.

kakkupikku wrote: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

The question is for some reason worded wierdly to me but I think the main point is that h = height in feet, t = time in seconds

The equation of height at t seconds is given as -16 (t - 3)^2 + 150

Now we want to find the height 2 seconds after it has reached its max height.

Max height will be reached when t = 3:

when t = 3, (-16*0) = 0 and you are not subtracting anything from the height of 150.

Thus when t = 3, h = 150 (MAX height)

2 sec after 3 is t = 5

so -16*(2)^2 +150

-64 + 150 = 86

(B)

than answer should be B
since (t-3)^2 is always positive ,so it will be minimum when t=3

after two second i.e 3=2=5 seconds
putting t=5 in the eq

h=-64+150
h=86

Very intuitive way to look at the problem. Good work. I see any sort of projectile motion and I instinctively think calculus

raleigh wrote:Very intuitive way to look at the problem. Good work. I see any sort of projectile motion and I instinctively think calculus

Lucky you..coz when i see time, distance, height, length, I immediately draw a line and start looking for who is going where :roll:

This is a gmat question. Was in set 26 of math Question #21.

Wow, great soln !!! I missed the "after reaching max height part" when i solved it.. thanks !!!