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?
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
Thank you Steve,
However, I still don't get your instructions for b) and c)
a) Since the supplies are dropped from a stationary helicopter, the initial velocity of the supplies is 0 feet per second. This means that v0 in the height equation is 0, simplifying the equation to:
y(t) = y0 - 16t^2
b) To determine when the supplies hit the ground, we need to find the value of t where y(t) = 0. Substituting y(t) from the height equation:
0 = y0 - 16t^2
Rearranging the equation:
16t^2 = y0
Solving for t:
t^2 = y0/16
Taking the square root of both sides:
t = √(y0/16)
So the supplies hit the ground at a time t = √(y0/16) seconds.
c) To find how high above the ground the helicopter initially was when the supplies were dropped, we need to find the value of y0. Given that the supplies hit the ground with a velocity of -160 feet per second, we can use this information to determine y0.
The velocity at time t is given by the derivative of the height function:
v(t) = y'(t) = -32t
At the time when the supplies hit the ground, t = √(y0/16) seconds. So plugging this value into the velocity function:
-160 = -32√(y0/16)
Simplifying the equation:
√(y0/16) = 160/32
√(y0/16) = 5
Squaring both sides:
y0/16 = 25
Multiplying both sides by 16:
y0 = 400
Therefore, the initial height of the supplies (and thus the helicopter) was 400 feet above the ground.