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?
Answer
a) Since the supplies are dropped from a stationary helicopter, their initial velocity is 0 feet per second. This simplifies the height equation because the term involving v0 becomes zero. So the height equation becomes:
y(t) = y0 - 16t^2
b) In order to find when the supplies hit the ground, we need to find the value of t when y(t) is zero. So we set the height equation to zero and solve for t:
0 = y0 - 16t^2
Rearranging the equation:
16t^2 = y0
Taking the square root of both sides:
t = sqrt(y0/16)
Note that we take the positive square root because time cannot be negative. Therefore, the supplies hit the ground at t = sqrt(y0/16).
c) To find the initial height of the helicopter (y0), we need to substitute the value of t in the original height equation:
y(t) = y0 - 16t^2
At t = sqrt(y0/16), we know that y(t) is zero (since the supplies hit the ground). Therefore, we can solve for y0:
0 = y0 - 16(sqrt(y0/16))^2
0 = y0 - 16(y0/16)
0 = y0 - y0
0 = 0
This indicates that there are infinitely many solutions for y0, meaning that the initial height of the helicopter could be any value.