I have a quadratic word problem, the question is (a ball is thrown up in the air. Its equation is
h= - 4.9t squared+38t+1.75 its asking for what length of time is the ball above 50m? Can someone show me how to do this. Thank you
2007-05-06 15:21:16 · 8 answers · asked by crocop49 1 in Science & Mathematics ➔ Mathematics
h = -4.9t² + 38t + 1.75
h = 50m, replacing
50 = -4.9t² + 38t + 1.75
4.9t² - 38t + 48.25 = 0..........using the quadratic formula
. . . . .-b±√(b²-4ac)
t = ----------------------, a=4.9, b = -38, c=48.25
. . . . . . .. .. 2a
. . . . .38±√(38²-4*4.9*48.25)
t = ----------------------------------
. . . . . . .. .. 2*4.9
. . . . .38±22.32
t = ------------------
. . . . . . .9.8
t = 1.6 sec
t = 6.155 sec
So, the ball passes the 50m at 1.6 seconds (going up) and at 6.155 sec (going down).
The time the ball is above 50m is: 6.155-1.6 = 4.555 seconds.
2007-05-06 15:34:47 · answer #1 · answered by zindagii_peru 4 · 0⤊ 0⤋
As others have said, use the quadratic formula:
h = -4.9t^2 + 38t + 1.75 = 50
<=> -4.9 t^2 + 38t - 38.25 = 0
<=> t = (-38 ± √(38^2 - 4(-4.9)(-38.25))) / (2(-4.9))
= 1.189, 6.566 (3 d.p.)
So it gets to 50m at 1.189 sec and falls below it again at 6.566 sec. So the length of time it is above 50m is 6.566 - 1.189 = 5.38 sec (2 d.p.)
2007-05-06 15:33:15 · answer #2 · answered by Scarlet Manuka 7 · 0⤊ 1⤋
in case you meant 2x/(3 - a million) = 7/x 2x/(3 - a million) = 7/x 2x×x = 7×2 2x^2 = 14 x^2 = 7 x = ±?7 so x = 2.64575 or -2.64575 Or did u recommend 2x/3 - a million = 7/x 2x/3 - a million = 7/x x(2x/3 - a million) = 7 2x^2/3 - x = 7 x^2 - 3x/2 = 21/2 (x - 3/4)^2 = 21/2 + 9/sixteen (x - 3/4)^2 = 168/sixteen + 9/sixteen (x - 3/4)^2 = 17716 x - 3/4 = ±?(177/sixteen) x = 3/4 ± ?(177/sixteen) so x = 4.07603 or -2.57603 desire this facilitates you. *??*
2016-10-30 12:46:16 · answer #3 · answered by arshad 4 · 0⤊ 0⤋
-4.9t^2+38t+1.75=50
solve for t
t =1.6 and 6.15 seconds approx.
2007-05-06 15:38:31 · answer #4 · answered by Scuba Dave 1 · 0⤊ 1⤋
sub in 50 for h
50=-4.9t^2+38t+175
0=-4.9t^2+38t+125
t=10.24
2007-05-06 15:25:18 · answer #5 · answered by leo 6 · 0⤊ 2⤋
There will be two times for which h = 50:
4.9t^2 - 38 t - 1.75 = - 50
4.9t^2 - 38 t + 48.25 = 0
t = ( 38 ± √(1444 - 945.7))/9.8
t = ( 38 ± √498.3)/9.8
t = ( 38 ± 22.323)/9.8
∆t = ( 38 + 22.323)/9.8 - ( 38 - 22.323)/9.8 = 2*22.323)/9.8 = 4.5556 sec.
2007-05-06 16:06:47 · answer #6 · answered by Helmut 7 · 0⤊ 0⤋
h = height, right?
You are given 50m as the height, right?
Set the entire equation to equal 50m and solve for t, time.
Guido
2007-05-06 15:28:23 · answer #7 · answered by Anonymous · 0⤊ 1⤋
h=50
50 = -4.9 t^2+38t+1.75
t = (-b +- √b^2-4ac)/2a
= [-38+-√38^2-4(-4.9)(-48.25)]2(-4.9)
values of t are
t = -1.6 and
t=6.155
therefore length of time is 6.155
2007-05-06 15:38:47 · answer #8 · answered by Maan_palaboy 2 · 0⤊ 1⤋