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
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