Asked by Anonymous
A bomb is dropped from a plane flying at constant horizontal velocity at an altitude of 600m. Where is the bomb after 2 sec? 5 sec? 10 sec? what are the velocities said period?
Answers
Answered by
Henry
h(t) = ho - 0.5g.t^2
h(2) = 600 - 4.9*2^2 = 580.4 m. Above gnd.
V(2) = Vo + g*t = 0 + 9.8*2 = 19.6 m/s.
h(5) = 600 - 4.9*5^2 = 477.5 m. Above gnd.
V(5) = 0 + 9.8*5 = 49 m/s.
h(10) = 600 - 4.9*100 = 110 m. Above gnd.
V(10) = 0 + 9.8*10 = 98 m/s.
h(2) = 600 - 4.9*2^2 = 580.4 m. Above gnd.
V(2) = Vo + g*t = 0 + 9.8*2 = 19.6 m/s.
h(5) = 600 - 4.9*5^2 = 477.5 m. Above gnd.
V(5) = 0 + 9.8*5 = 49 m/s.
h(10) = 600 - 4.9*100 = 110 m. Above gnd.
V(10) = 0 + 9.8*10 = 98 m/s.
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