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?
Neglect air friction.
Solve this equation for height, H:
Y (height above ground)
= H + 20t -4.9 t^2 = 0
using t = 15 s.
After leaving the helicopter the package is moving upwards at the initial velocity v0.
This is decelerated motion with acceleration –g (g≈10m/s^2)
h0=v0t1-(gt^2)/2
v=0=v0-gt1
t1=v0/g=20/10=2 s.
After this time interval the package is at the height
h0=v0^2/2g= 20 m.
Then the package is falling down during t2=t-t0=15-2=13 s:
H= h0+h=(gt2^2)/2=(10x169)/2=845 m.
Hence, the height of helicopter when the package was dropped is
h=H-h0=845-20=825 m
To determine the height of the helicopter when the package was dropped, we can use the formula for height:
h = v*t,
Where h is the height, v is the velocity, and t is the time taken.
Given that the velocity is 20 m/s and the time taken is 15 seconds, we can now calculate the height:
h = 20 m/s * 15 s,
h = 300 m.
Therefore, the helicopter was approximately 300 meters high when the package was dropped, to the nearest meter.
To determine the height of the helicopter when the package was dropped, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = height (unknown)
u = initial velocity (20 m/s upwards)
t = time taken (15 seconds)
a = acceleration (acceleration due to gravity, approximately 9.81 m/s² downward)
Since the helicopter is traveling upwards, the initial velocity (u) would be considered negative.
Substituting the given values into the equation, we get:
s = (-20 m/s) * 15 s + (1/2) * (-9.81 m/s²)(15 s)^2
Simplifying further:
s = -300 m - (1/2) * 9.81 m/s² * 225 s²
s = -300 m - 1102.625 m
s ≈ -1402.625 m
Since the height cannot be negative, we take the absolute value of the result to get the height of the helicopter when the package was dropped.
The height of the helicopter when the package was dropped is approximately 1403 meters.