h(t) = 16t^2 - 40t + 50 represents the height, in feet, of an object above level ground at time t, in seconds. Given this information, complete the following sentence with one of the choices listed below.
The closest to the ground that the object gets at any time t > 0 is ____ feet.
50 , 40 , 32 , or 25
2006-10-14 06:18:58 · 10 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
Find the minimum value of h(t).
h'(t) = 32t - 40. h'(t) = 0 when t=5/4.
h''(t) = 32, so h(t) > 0 for all t. Therefore the critical point t=5/4 corresponds to a minimum.
To get the minimum height, plug in t=5/4 and you get
16*(25/16) - 40(5/4) + 50 = 25 - 50 + 50 = 25.
2006-10-14 06:28:09 · answer #1 · answered by James L 5 · 0⤊ 0⤋
The answer is 25. That's the minimum point in the graph using the given expression.
first find the derivative of h(t)=16t^2-40t+50
h'(t)=32t-40
To get the lowest point equate the derivative to 0.
32t-40=0
t=40/32 = 1.25 sec.
Now substitute 1.25 sec in the given expression:
16(1.25)^2-40(1.25)+50=25
This is a problem in differential calculus. You remember the maxima and minima section? Get the derivative and equate it to 0 to solve for t.
You asked a similar question sometime back. The graph is a parabola with its lowest point at 25 feet. Sorry I gave you a wrong answer in that other question of yours. tulb
2006-10-14 17:07:01 · answer #2 · answered by srgolfer 1 · 0⤊ 0⤋
If t=0, the equation yields 50.
( h(0)=16(0)^2 - 40(0) + 50 = 50)
Hence, with t > 0, the closest the object would be to the ground would be 50 ft.
2006-10-14 13:23:52 · answer #3 · answered by Rie 3 · 0⤊ 1⤋
There are a couple of ways to do this. The first is just to graph it. The more formal way is to take the first derivative and set it to zero. When you find the time where the first derivative is zero, plug it back in to find the answer. My guess looking at the choices is that you should graph it. Hint: Look that the height at 1 second.
2006-10-14 13:23:46 · answer #4 · answered by Anonymous · 0⤊ 0⤋
50 feet
2006-10-14 13:29:42 · answer #5 · answered by openpsychy 6 · 0⤊ 0⤋
in physics your question is equivalent to x = x (0) + V0T + 1/2aT*2
So the initial height X(0) the object is dropped in your equation is 50
2006-10-14 13:29:17 · answer #6 · answered by mystic_golfer 3 · 0⤊ 0⤋
25 - complete the square and then everything becomes clear...
2006-10-14 13:29:08 · answer #7 · answered by didyourmumnottellyoustaringsrude 3 · 0⤊ 0⤋
i say 25 feet at t=1.25 seconds...i graphed it
2006-10-14 13:29:00 · answer #8 · answered by cheesy 2 · 0⤊ 0⤋
I'm not doing your homework!
2006-10-14 13:26:15 · answer #9 · answered by JennyfferBCN 5 · 0⤊ 1⤋
Do your homework alone.....
2006-10-14 13:26:56 · answer #10 · answered by George 2 · 0⤊ 1⤋