A bomber flies horizontally with a speed of

367 m/s relative to the ground. The altitude
of the bomber is 4020 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?
Answer in units of m

(bomber speed)*(bomb fall time)

To find the distance from the point vertically under the point of release where the bomb hits the ground, we can use the equations of motion.

First, we need to determine the time it takes for the bomb to hit the ground. Since the bomb is released horizontally and there is no horizontal acceleration, the time it takes to hit the ground is determined by the vertical motion. We can use the equation of motion for vertical displacement:

h = ut + (1/2) gt^2

where h is the initial vertical height (altitude), u is the initial vertical velocity (which is zero in this case as the bomb is released vertically), g is the acceleration due to gravity, and t is the time.

Plugging in the values:
h = 4020 m
u = 0 m/s
g = 9.8 m/s^2

We can rearrange the equation and solve for time:

(1/2) gt^2 = h

(1/2) * 9.8 * t^2 = 4020

4.9 * t^2 = 4020

Divide both sides of the equation by 4.9:

t^2 = 4020 / 4.9

t^2 ≈ 821.63

Taking the square root of both sides to solve for t:

t ≈ √(821.63)

t ≈ 28.65 seconds

Now, we know that the distance covered horizontally in this time is given by:

d = v*t

where v is the horizontal velocity of the bomber and d is the distance.

Plugging in the values:
v = 367 m/s
t = 28.65 s

d = 367 * 28.65

d ≈ 10539.55 m

Therefore, the bomb hits the ground approximately 10539.55 m away from the point vertically under the point of release.