A bomber flies horizontally with a speed of
231 m/s relative to the ground. The altitude
of the bomber is 2520 m and the terrain is
level. Neglect the effects of air resistance.
The acceleration of gravity is 9.8 m/s
a) How far from the point vertically under the point of release does a bomb hit the ground?
Answer in units of m.
How long does it take to hit the ground?
4.9t^2 = 2520
t = 22.67 s
The horizontal speed of the bomb is constant -- that of the plane when it was released. As usual, distance = speed * time, so the bomb will hit
231 * 22.67 = 5238.6 m downrange
To find the distance from the point vertically under the point of release where the bomb hits the ground, we can use the equation of motion:
s = ut + 0.5 * a * t^2
Where:
s = distance traveled
u = initial velocity
a = acceleration (gravity)
t = time
At the point of release, the vertical velocity of the bomb is 0 m/s since it is purely horizontal. The initial altitude of the bomber is 2520 m, so the time taken for the bomb to reach the ground can be found using the equation:
s = ut + 0.5 * a * t^2
0 = 0 * t + 0.5 * (-9.8) * t^2
Solving this equation will give us the time it takes for the bomb to hit the ground.
0 = -4.9 * t^2
t^2 = 0
Since the time squared is zero, it means the bomb hits the ground instantly at the point vertically under the point of release. So, the distance from the point vertically under the point of release where the bomb hits the ground is 0 meters.
To find the distance from the point vertically under the point of release to where the bomb hits the ground, we can use the concept of horizontal motion and vertical motion independently.
First, let's calculate the time it takes for the bomb to hit the ground.
Using the equation of motion for vertical motion:
h = (1/2) * g * t^2
Where h is the initial height (altitude) and g is the acceleration due to gravity.
Rearranging the equation:
t^2 = (2 * h) / g
Substituting the given values:
t^2 = (2 * 2520 m) / 9.8 m/s^2
t^2 ≈ 514.29
t ≈ √514.29
t ≈ 22.69 seconds
Now, let's calculate the horizontal distance traveled by the bomb during this time.
Using the formula for horizontal distance:
d = v * t
Where d is the horizontal distance, v is the horizontal velocity, and t is the time of flight.
Substituting the given values:
d = 231 m/s * 22.69 s
d ≈ 5236.39 meters
Therefore, the bomb hits the ground approximately 5236.39 meters from the point directly beneath its release point.
Answer: The bomb hits the ground approximately 5236.39 meters from the point vertically under its release point.