A bomber flies horizontally with a speed of 431 m/s relative to the ground. The altitude of the bomber is 2630 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) To find the horizontal distance that the bomb travels, we need to calculate the time it takes to reach the ground using the given altitude and the acceleration due to gravity.
The equation to calculate the time taken is:
h = (1/2) * g * t^2
where h is the altitude, g is the acceleration due to gravity, and t is the time.
Substituting the given values:
2630 = (1/2) * 9.8 * t^2
Rearranging the equation to solve for time:
t^2 = (2 * 2630) / 9.8
t^2 = 536 / 9.8
t^2 ≈ 54.69
t ≈ √54.69
t ≈ 7.4 seconds (approximately)
Now, we can calculate the horizontal distance using the time and the horizontal speed of the bomber.
distance = speed * time
distance ≈ 431 * 7.4
distance ≈ 3187.4 m
Therefore, the bomb hits the ground approximately 3187.4 m from the point directly below the point of release.
B) To find the angle at the point of release, we can use trigonometry.
The vertical component of the speed is the rate of descent of the bomb, which can be calculated using the time calculated earlier:
vertical speed = altitude / time
vertical speed ≈ 2630 / 7.4
vertical speed ≈ 355.14 m/s
The horizontal component of the speed is the same as the horizontal speed of the bomber, which is given as 431 m/s.
The tangent of the angle is given by:
tan(angle) = vertical speed / horizontal speed
Substituting the values:
tan(angle) ≈ 355.14 / 431
angle ≈ arctan(355.14 / 431)
angle ≈ 39.94 degrees (approximately)
Therefore, the telescopic bomb sight must be set at an angle of approximately 39.94 degrees from the vertical at the point of release.
To solve this problem, we need to break it down into two parts - horizontal motion and vertical motion.
A) How far from the point vertically under the point of release does a bomb hit the ground?
We know that the horizontal speed of the bomber is 431 m/s, and the altitude (vertical distance) of the bomber is 2630 m. Since we need to find how far the bomb lands horizontally from the point vertically under the release point, we need to calculate the horizontal distance traveled by the bomb.
To do this, we can use the formula for horizontal distance: distance = speed × time. In this case, the time will be the same for both the bomb and the bomber, because they are dropped at the same time. Therefore, we can use the time it takes for the bomber to reach the ground.
To find the time it takes for the bomber to reach the ground, we can use the formula for vertical motion: distance = initial velocity × time + (1/2) × acceleration × time^2.
Given:
- Initial vertical velocity of the bomber = 0 m/s (since it is dropped vertically)
- Final vertical velocity of the bomber = ? (since we need to find the time it takes for the bomb to hit the ground)
- Acceleration due to gravity = 9.8 m/s^2
- Vertical distance traveled by the bomber = 2630 m
Using the formula, we can rearrange to solve for time:
2630 = (1/2) × 9.8 × time^2
Divide both sides by (1/2) × 9.8:
time^2 = (2 × 2630) / 9.8
time^2 = 536.73
Take the square root of both sides:
time = √536.73 = 23.18 seconds (approximately)
Now that we have the time, we can calculate the horizontal distance traveled by the bomber:
distance = speed × time
distance = 431 × 23.18
distance ≈ 9987.58 meters
Therefore, the bomb will hit the ground approximately 9987.58 meters from the point vertically under the release point.
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?
To find the angle, we need to determine the velocity of the bomb at the moment of release. The horizontal component of the bomb's velocity will be the same as the bomber's speed (431 m/s), and the vertical component will be determined by the time calculated in part A.
Using the formula for vertical motion, the vertical component of the bomb's velocity can be calculated:
final vertical velocity = initial vertical velocity + acceleration × time
final vertical velocity = 0 + 9.8 × 23.18
final vertical velocity ≈ 227.12 m/s
Now, we can find the angle using trigonometry. The tangent of the angle is equal to the vertical component divided by the horizontal component:
tan(angle) = (vertical component) / (horizontal component)
tan(angle) = 227.12 / 431
angle ≈ 29.93 degrees
Therefore, the telescopic bomb sight must be set at an angle of approximately 29.93 degrees from the vertical at the point of release for the bomb to hit the target seen in the sight.
A. h = Vo*t + 0.5g*t^2 = 2630 m.
0 + 4.9t^2 = 2630
t^2 = 536.73
Tf = 23.2 s. = Fall time or time in air.
d = Xo * Tf = 431m/s * 23.2s. = 9985 m.
B. tanA = Y/X = 2630/9985 = 0.26340
A = 14.8o