Posted by Martin on Tuesday, April 16, 2013 at 1:17am.
Supplies are dropped from a stationary (not moving up or down) helicopter and, after some amount of time, land on the ground with a velocity of 160 feet per second.
Assumptions: Start time at t = 0 and ignore air resistance. Recall that the height function is given by
y(t)= y0 + v0t  16t^2
where y0 is the initial height of the supplies (in feet) and v0 is the initial
velocity of the supplies (in feet per second).
a) Since the supplies are dropped from a
stationary helicopter, what is the initial velocity of the supplies? How does this simplify the height equation?
b) Using the fact that the supplies hit the ground with a velocity of 160 feet per second, when do the supplies hit the ground?
c) Using previous parts, how high above the ground was the helicopter initially when the supplies were dropped?

Math/ Calculus Help1  Steve, Tuesday, April 16, 2013 at 11:43am
a) v0=0, so y(t) = y016t^2
b) v(t) = 32t, so solve for t when v=160
c) now plug that t into y(t)=0 to get y0

Math/ Calculus Help1  Angela, Tuesday, April 16, 2013 at 10:29pm
Thank you Steve,
However, I still don't get your instructions for b) and c)
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