Answer:8730.63 m
Explanation:
Given
Bomber velocity [tex]u=323 m/s[/tex] relative to ground
Altitude of the bomber [tex]h=3580 m[/tex]
Range of bomb when bomb when it is released is given by
[tex]R=u\times t[/tex]
where [tex]u =velocity\ of\ Bomber[/tex]
Horizontal velocity of bomber is equal to bomb horizontal velocity
time taken by to cover 3580 m
[tex]h=ut+\frac{at^2}{2}[/tex]
[tex]3580=0+\frac{9.8\times t^2}{2}[/tex]
[tex]t^2=\frac{7160}{9.8}[/tex]
[tex]t=27.03 s[/tex]
Therefore range of bomb is
[tex]R=323\times 27.03=8730.63 m[/tex]
Therefore bomber must release the bomb at a horizontal distance of 8730.63 m before the hit point
