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28 Oct 2025, 11:01
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Question Stats:
100% (01:05) correct
0% (00:00) wrong based on 2 sessions
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The height, h, of a thrown ball as a function of t, the amount of time it has been in the air, is given by h(t) = –10t^2 + 40t. What is the maximum height attained by the ball?
A. 22
B. 0
C. 40
D. 80
E. 160
A. 22
B. 0
C. 40
D. 80
E. 160
Updated on: 02 Nov 2025, 10:01
Originally posted by Sajjad1994 on 02 Nov 2025, 09:54.
Last edited by Sajjad1994 on 02 Nov 2025, 10:01, edited 1 time in total.
Last edited by Sajjad1994 on 02 Nov 2025, 10:01, edited 1 time in total.
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Official Explanation
Factor out the common factor, \(–10t,\) to obtain \(h(t) = –10 t(t – 4)\). Find the t-intercepts (x-axis intercepts) of the graph by setting \(y = h(t) = 0.\) Then \(0 = –10 t(t – 4).\) The product is 0 when either factor is 0, so \(–10t = 0\) or \(t – 4 = 0,\) and \(t = 0\) or \(t = 4.\) By the symmetry of a parabola, the vertex will lie on the vertical line t = 2, because 2 is halfway between the t-intercepts of 0 and 4. The maximum height of the ball will be the y-coordinate of the vertex, namely h(2), which is by definition the height of the ball at time 2. h(2) = –10 × 22 + 40 × 2 = –40 + 80 = 40, (C).
Answer: C
GMAT-Club-Forum-lmf35baz.png [ 55.77 KiB | Viewed 64 times ]
Factor out the common factor, \(–10t,\) to obtain \(h(t) = –10 t(t – 4)\). Find the t-intercepts (x-axis intercepts) of the graph by setting \(y = h(t) = 0.\) Then \(0 = –10 t(t – 4).\) The product is 0 when either factor is 0, so \(–10t = 0\) or \(t – 4 = 0,\) and \(t = 0\) or \(t = 4.\) By the symmetry of a parabola, the vertex will lie on the vertical line t = 2, because 2 is halfway between the t-intercepts of 0 and 4. The maximum height of the ball will be the y-coordinate of the vertex, namely h(2), which is by definition the height of the ball at time 2. h(2) = –10 × 22 + 40 × 2 = –40 + 80 = 40, (C).
Answer: C
Attachment:
GMAT-Club-Forum-lmf35baz.png [ 55.77 KiB | Viewed 64 times ]
