Neglect air friction.
Solve this equation for height, H:
Y (height above ground)
= H + 20t -4.9 t^2 = 0
using t = 15 s.
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?
2 answers
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
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