A bomber flies horizontally with a speed of 235 m/s relative to the ground. The altitude of the bomber is 3570 m and the terrain is level. Neglect the effects of air resistance.
The acceleration of gravity is 9.8 m/s2 .
a) How far from the point vertically un- der the point of release does a bomb hit the ground?
To find the horizontal distance the bomb travels before hitting the ground, we need to calculate the time it takes for the bomb to fall vertically. We can use the equation of motion for vertical motion:
h = (1/2) * g * t^2
where
h = altitude (3570 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation to solve for time (t), we get:
t^2 = (2h) / g
t = √[(2h) / g]
t = √[(2 * 3570) / 9.8]
t ≈ 8.49 seconds
Now that we know the time it takes for the bomb to fall, we can calculate the horizontal distance (D) it travels using the equation:
D = v * t
where
v = horizontal velocity of the bomber (235 m/s)
t = time (8.49 seconds)
Plugging in the values, we have:
D = 235 * 8.49
D ≈ 1992.15 meters
Therefore, the bomb hits the ground approximately 1992.15 meters away from the point vertically beneath the point of release.