S=(1/2)gt^2 since u=0
4000=(1/2)*10*t^2
t=sqrt(800)secs - time to reach ground
So hor. distance covered= 80*sqrt(800)
= 2263m
A bomb is dropped from a plane flying at 4000 meters. The plane is flying at 80 m/s. How far, in a horizontal direction, did the bomb hit the ground?
2 answers
h = ho - o.5g*t^2 = 0
4000 - 4.9t^2 = 0
-4.9t^2 = -4000
t^2 = 816.3
Tf = 28.6 s. = Fall time or time in flight.
Dx = Hor = 80m/s * 28.6s = 2286 m.
4000 - 4.9t^2 = 0
-4.9t^2 = -4000
t^2 = 816.3
Tf = 28.6 s. = Fall time or time in flight.
Dx = Hor = 80m/s * 28.6s = 2286 m.