I need help solving this problem. Please show all work. A ball moves in a vertic

Question

I need help solving this problem. Please show all work.

A ball moves in a vertical air tunnel in response to air flow in the tunnel. Suppose That y(t) represents the vertical position of the center of the ball above 'ground level'. (Note: It is possible for the ball to be below ground level.) Air is fed into the tunnel at varying rates beginning at t = - 2 and ending at t = 14. Assume that it is known that y(0) = 10 and that the vertical velocity of the center of the ball is given by v(t) = Find y(2) and y(8). From t = 0 to t = 13, at what instant of time does the ball reach the lowest point in its trajectory? At what instant does the ball reach the highest point in its trajectory?

Explanation / Answer

for t=2 sec,

dy/dt = t^2 -t -6 (for t<4)

y = t^3/3 -t^2/2 -6t +c

putting t = 0, y(0) = 10

=> c = 10

y(2) = -1.33

y(4) = -0.66

y(t>4) = 8.667t -t^2/3 + c

y(t=4) = -0.66

=> c = -30

y(8) = 18

b)

lowest point till t = 4,

v = 0 at t=3,

y = -3.5 => lowest point till t = 4

highest point = 10 at t=0

from 4 to 13,

v = 0 at t=13

y(13) = 26.33, which is the highest point.

lowest will be t=4 which is -0.66

Combining,

highest = 26.33 at t=13

lowest -3.5 at t=3

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