A bomber is flying at a velocity of 200.0 km/hr with an altitude of 1.5 km. How far (horizontally) from its intended target should the plane drop the bomb?

Answer 1

The bomb moves exactly as fast horizontally as the bomber if friction is ignored. (If the bomber does not turn, the bomb explodes right under it)

So how long will it take the bom to fall 1500 meters?
h = (1/2) g t^2
1500 = 4.9 t^2
t = 17.5 seconds to fall.
how far does the bomber move in 17.5 seconds?
(200,000 m/3600 s)(17.5s) = 972 meters

Answer 2

A launcher fires a projectile with a velocity of 6.31m/s. Determine the horizontal distance of the projectile (in the air) at 5 degrees. Recalculate this distance at 10, 15, 20.... (5 degree increments) up to 90 degrees (at which point the horizontal distance will be zero). Graph this data (Distance vs. Launch Angle). Distance will be on the Y-axis and launch angle will be on the X-axis.