Consumption and Real Estate
A package of supplies is to be dropped from an airplane so that it hits the ground at a designated spot near?
some campers. The airplane, moving horizontally at a constant velocity of 140 km/h, approaches the spot at an altitude of 0. 500 km above level ground. Having the designated point in sight, the pilot prepares to drop the package. (a) What should the angle be between the horizontal and the pilot’s line of sight when the package is released? (b) What is the location of the plane when the hits the ground?
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#2 written by anonimous 1 year ago
You have not described enough parameters to answer these questions. The mechanism that releases the package of supplies may or may not impart a velocity to the package. Is there a parachute? What is the terminal velocity of the package whether or not there is a parachute. Dropping half a kilometer would do a lot of damage upon impact to something that had a high terminal velocity.
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#3 written by datuk M 1 year ago
first of all, you resolve the velocity of the plane into x and y component. then use formula v^2=u^2-2gs where v is final velocity which is 0 and u initial velocity usin(?) .
make sure that the altitude is negative sign because it is in y component and pointind downwards.put all the values given and you can determine the value of (?). - Comment Feed for this Post
V is the velocity of the plane and ө is the angle of sight of the spot from the horizontal.
The height from the ground is H.
The horizontal distance on the ground of the spot from the vertical from the plane at that instant is S.
Then tanө = H/S————–(a).
H =g T^2(1/2)
T^2 = 2H/g ——- (b)
S = VT
S^2 = V^2*T^2
S^2 = V^2* (2H/g) from (b)
(H/tanө)^2 = V^2* (2H/g) from (a)
H^2 = V^2* (2H/g) (tanө)^2
(tanө)^2 = gH/ (2 V^2)
g = 9.8 m/s^2, H =500 x10^3 m and V= 38.9 m/s
tanө = 40.24
ө = 88.6 degree.
b) Vertically above the spot at an altitude of 500 km.