# Math/ Calculus Help1

posted by on .

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 - ,

a) v0=0, so y(t) = y0-16t^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 - ,

Thank you Steve,

However, I still don't get your instructions for b) and c)