Time for 100m. drop = sqrt. (2h/g) = sqrt.(200/9.8) = 4.52 secs.
(200/3.6) = 55.56m/sec.
Distance before target for release = (55.56 x 4.52) = 251.13 metres.
Sight angle below horizontal = arctan (100/2571.13) = 21.7 degrees.
The pilot of an airplane carrying a package of mail to a remote outpost wishes to release the package at the right moment...?
The pilot of an airplane carrying a package of mail to a remote outpost wishes to release the package at the right moment to hit the recovery location "A." What angle θ with the horizontal should the pilot's line of sight to the target make at the instant of release? The airplane is flying horizontally at an altitude of 100 m with a velocity of 200 km/ h.
- falling time t = √2h/g = √2*100/9.806 = 4.516 sec
horizontal distance d = V*t = 200*4.516 /3.6 = 250.9 m
Θ angle = arctan -h/d = arctan (-100/250.9) = -21.73°10
- 200 km/hr (1000 m/km / 3600 s/hr) = 55.6 m/s
neglect air resistance, assume g = 9.8 m/s²
The time needed to fall 100 m from vertical rest is
s = ½at²
t = √(2s/a)
t = √(2(100) / 9.8)
t = 4.52 s
the distance the plane travels in 4.52 s is
s = vt
s = 55.6(5.42)
s = 251 m
tanθ = 100 / 251
θ = 21.7° below the horizon
I hope this helps.
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