Difficult Math Problem(regarding parabolas)?

A ball is thrust up vertically from the ground into the air and hits the ground 2.5 seconds

later. What is the maximum height of the ball in feet?. Assume that air resistance is

negligible. The acceleration due to gravity is -32 ft/sec2.

A: 50

B: 30

C: 100

D: 75

E: none of the above

Comments

  • total time = 2.5 sec

    time up + time down = total time

    time up = time down

    time down = 1.25 sec

    acceleration = -32 ft/sec^2

    initial velocity = 0 ft/sec

    final velocity = -40 ft/sec

    average velocity = -20 ft/sec

    maximum height = (time)(avg velocity)(-1)

    = (1.25 sec)(-20 ft/sec)(-1)

    = 25 ft

    Therefore, the maximum height of the ball is 25 feet.

  • All element on x-axis have coordinates (x, 0). to discover the factors on the x-axis that are additionally on the parabola, basically discover each and all of the factors on the parabola with coordinates (x, 0) y = 6x - 4x² If y = 0, then 0 = 6x - 4x² = 2x(3 - 2x) this is the two x= 0 or x = 3/2 The coordinates of the factors the place the parabola crosses the x-axis are (0, 0), (3/2, 0)

  • If total time is 2.5 secs then time from ground to highest point is 1.25 seconds.

    v = u-gt

    0 = u - 32*1.25

    so initial speed is 40 ft/ sec

    substitute in x = ut - (0.5)g(t^2)

    = (40*1.25) - (0.5)(32)(1.25^2)

    = 25 so answer is E

  • h=g*t^2/2, t=2,5/2=5/4, h=32*(5/4)^2/2=25ft

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