A bomber flies horizontally with a speed of

225 m/s relative to the ground. The altitude
of the bomber is 3010 m and the terrain is
level. Neglect the effects of air resistance.
The acceleration of gravity is 9.8 m/s
2
.
a) How far from the point vertically under the point of release does a bomb hit the
ground?
Answer in units of m

b) At what angle from the vertical at the point
of release must the telescopic bomb sight be
set so that the bomb hits the target seen in
the sight at the time of release?
Answer in units of

a. find how long it takes an object to free fall 3010 m. then, horizontal distance=225m/s*time

b. what is
arc tan(3010/horizontaldistance) ?

To solve this problem, we can use the equations of motion in two dimensions. Let's take the positive direction as upwards.

a) First, let's find the time it takes for the bomb to hit the ground. We can use the equation:

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

where h is the initial altitude, u is the initial vertical velocity (0 m/s since the bomber is flying horizontally), g is the acceleration due to gravity, and t is the time.

Substituting the given values, we have:

3010 = 0*t + (1/2)(9.8)t^2

Rearranging the equation, we get:

4.9t^2 = 3010

Solving for t, we find:

t^2 = 3010 / 4.9
t^2 = 614.29
t ≈ 24.78 s

Now, let's find the horizontal distance the bomb travels. We can use the equation:

s = ut + (1/2)at^2

where s is the horizontal distance, u is the initial horizontal velocity (225 m/s), a is the horizontal acceleration (0 m/s^2), and t is the time.

Substituting the given values, we have:

s = 225 * 24.78 + (1/2) * 0 * (24.78)^2
s = 5569.5 m

b) To find the angle at which the bomb sight should be set, we need to calculate the horizontal and vertical components of the velocity at the time of release.

The horizontal component of the velocity remains constant throughout the motion and is equal to the initial horizontal velocity (225 m/s).

The vertical component of the velocity can be found using the equation:

v = u + at

where v is the final vertical velocity, u is the initial vertical velocity (0 m/s), a is the acceleration due to gravity, and t is the time.

Substituting the given values, we have:

v = 0 + 9.8 * 24.78
v ≈ 242.04 m/s

Now, we can calculate the angle θ using the equation:

θ = arctan(v/horizontal_velocity)

θ = arctan(242.04 / 225)
θ ≈ 47.74°

So, the bomb sight should be set at an angle of approximately 47.74° from the vertical at the point of release to hit the target.