Last visit was: 21 Nov 2024, 10:53 It is currently 21 Nov 2024, 10:53

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 30001
Own Kudos [?]: 36335 [3]
Given Kudos: 25926
Send PM
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [4]
Given Kudos: 136
Send PM
Verbal Expert
Joined: 18 Apr 2015
Posts: 30001
Own Kudos [?]: 36335 [0]
Given Kudos: 25926
Send PM
avatar
Manager
Manager
Joined: 07 Aug 2016
Posts: 59
Own Kudos [?]: 68 [2]
Given Kudos: 0
GRE 1: Q166 V156
Send PM
Re: An object thrown directly upward is at a height of h feet af [#permalink]
2
Given -16 (t-3)^2 + 150 and that the question is asking for maximum height, the maximum height occurs at t = 3 to zero out the - 16 (t - 3)^2

Maximum height = - 16 (3-3)^2 + 150 = 150

Now what happens after it reaches maximum height and continues for another 2 seconds, in addition to the 3 seconds to reach maximum height?

-16 (5-3)^2 + 150 = -16 (2^2) + 150 = - 16 * 4 + 150 = - 64 + 150 = 86
avatar
Intern
Intern
Joined: 24 Jan 2020
Posts: 24
Own Kudos [?]: 35 [0]
Given Kudos: 0
Send PM
Re: An object thrown directly upward is at a height of h feet af [#permalink]
1
Equation for height of object h = 150 - 16\((t-3)^2\)

The maximum height h the object can travel will occur only when the term 16\((t-3)^2\) is zero

Hence t-3 = 0 which makes t=3

Maximum height is reached when t=3

So the time which is 2 secs after maximum height = 3 + 2 = 5

Height at 5 secs = 150 - 16\((5 - 3)^2\) = 86

Answer is B
avatar
Intern
Intern
Joined: 07 Apr 2020
Posts: 6
Own Kudos [?]: 14 [0]
Given Kudos: 0
Send PM
Re: An object thrown directly upward is at a height of h feet af [#permalink]
1
The question can be rewritten as: h = 150 - 16 (t-3)^2

As (t-3)^2 always greater than 0, so -16(t-3)^2 =< 0, so 150 - 16 (t-3)^2 =< 150, so the maximum height will be 150 when (t-3)^2 = 0, which means t = 3

The question asked at the time of 2 seconds after the max height, therefore t'= 3+2 = 5. Apply to the equation, we have:

h = 150 - 16 (5-3)^2 = 150 - 16* (2^2) = 150 - 16*4 = 150-64 = 86

Therefore, the answer is B
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5030
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: An object thrown directly upward is at a height of h feet af [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
Re: An object thrown directly upward is at a height of h feet af [#permalink]
Moderators:
GRE Instructor
83 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne