A passenger in a helicopter traveling upwards at 20 m/s accidentally drops a package out the window. If it takes 15 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?
h = Vo*t + 0.5g*t^2.
h = -20*15 + 4.9*(15)^2 = 803 m.
To find the height of the helicopter when the package was dropped, we can use the kinematic equation:
h = h0 + v0*t + (1/2)*a*t^2
Where:
h = final height of the helicopter (what we need to find)
h0 = initial height of the helicopter (unknown, but what we need to find)
v0 = initial velocity of the helicopter (20 m/s upwards)
t = time taken to reach the ground (15 seconds)
a = acceleration (in this case, acceleration due to gravity, which is positive and equal to 9.8 m/s^2, acting downwards)
We need to assume that the velocity of the package is negligible since it is dropped, and in this case, we are neglecting the air resistance.
Since the helicopter is traveling upwards, the initial velocity (v0) will be negative. So, the equation becomes:
h = h0 - 20*t + (1/2)*9.8*t^2
Now, we can substitute the given values into the equation:
h = h0 - 20*15 + (1/2)*9.8*(15^2)
h = h0 - 300 + 7.35*(225)
h = h0 - 300 + 1653.75
Simplifying the equation, we get:
h = h0 + 1353.75
Since the question asks for the height to the nearest meter, we can round the answer to the nearest whole number, disregarding the decimal part.
Therefore, the helicopter was approximately 1354 meters above the ground when the package was dropped.