d = Vo*t + 0.5g*t^2 = 10.000 Ft.
0 + 4.9t^2 = 10,000,
t^2 = 2041,
t = Tf = 45.2 s. = Fall time = Time in
flight.
Dx = Xo * Tf = 300 m/s * 45.2 s. = 13,553 Ft.= Dist. it should be released from target.
a bomb is dropped from an airplane travelling horizontally with a speed of 300 mi/hr if the airplane is 10000ft. above the ground, how far from the target must it be released, neglect friction?
4 answers
Correction:
d = Vo*t + 0.5g*t^2 = 10,000 Ft.
0 + 16t^2 = 10,000,
t^2 = 625,
t = 25 s.
X0=300 mi/h * 1h/3600 s. = .0833 mi/s.
Dx = 0.08333 mi/s * 25s = 2.083 mi =
3333 m.
d = Vo*t + 0.5g*t^2 = 10,000 Ft.
0 + 16t^2 = 10,000,
t^2 = 625,
t = 25 s.
X0=300 mi/h * 1h/3600 s. = .0833 mi/s.
Dx = 0.08333 mi/s * 25s = 2.083 mi =
3333 m.
Where did you get the 16?
PURSIGIDO
1 answer