An airplane traveling 1001 m above the ocean at 125km/h is going to drop a box of supplies to shipwrecked victims below. What is the horizontal distance between the plane and the victims when the box is dropped?
5 answers
The time is 14.3s, btw..
The time T that it will take the box to fall 1001 m, neglectring air resistance, is given by the equation
(g/2)T^2 = 1001
T^2 = 2002/9.8 = 204.3 s^2
T = 14.3 s
The distance that the box moves forward while it falls is V*T.
You will need to convert the 125 km/h speed to ___ m/s to get the distance in meters.
(g/2)T^2 = 1001
T^2 = 2002/9.8 = 204.3 s^2
T = 14.3 s
The distance that the box moves forward while it falls is V*T.
You will need to convert the 125 km/h speed to ___ m/s to get the distance in meters.
Givens:
a=-9.81m/s^2
d=-1001m
vi=0m/s
d=vit+1/2at^2
-1001m=1/2-9.81m/s^2t^2
-1001m=-4.905m/s^2t^2
divide -1001m by -4.905m/s^2
204s^2=t^2
square root both sides
14.3s=t
125km/hr is 56m/s
56*14.3=
800.8m, that's the distance from the package dropped.
a=-9.81m/s^2
d=-1001m
vi=0m/s
d=vit+1/2at^2
-1001m=1/2-9.81m/s^2t^2
-1001m=-4.905m/s^2t^2
divide -1001m by -4.905m/s^2
204s^2=t^2
square root both sides
14.3s=t
125km/hr is 56m/s
56*14.3=
800.8m, that's the distance from the package dropped.
Woops!
125km/hr is 34.7m/s
so 34.7*14.3 is 496.21m
125km/hr is 34.7m/s
so 34.7*14.3 is 496.21m
7.8m/s