How Do You Calculate the Impact Point of a Bomb Released from a Moving Airplane?

[33639]

I'm having problems with this question. I don't know how to solve it...I've usually solved questions with some sort of resistance force but this I don't know how to solve.
How would I solve this question?

1. The problem

A bomber flies horizontally with a speed of 275 m/s relative to the ground. The altitude of the bomber is 3000. m and the terrain is level. Neglect the effects of air resistance.
(a) How far from the point vertically under the point of release does a bomb hit the ground?
(b) If the plane maintains its original course and speed, where is it when the bomb hits the ground?
(c) 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 t he sight at the time of release?

 

[ehild]

Think: at the instant when the bomb is released it has the same horizontal velocity as the bomber. So you have a falling body with horizontal initial velocity: it is a projectile. You know the laws of projectile motion?

ehild

 

[33639]

Think: at the instant when the bomb is released it has the same horizontal velocity as the bomber. So you have a falling body with horizontal initial velocity: it is a projectile. You know the laws of projectile motion?

ehild

I'm just confused with how can I find the speed of the bullet as it's traveling down.
I know in the x component v = 275 and in the y component d = 3000 and a = -9.8

 

[HallsofIvy]

This is very strange. Problems with "some sort of resistance force" are usually much harder that problems like this.

"A bomber flies horizontally with a speed of 275 m/s relative to the ground. The altitude of the bomber is 3000. m"
There is NO horizontal force on the bullet and so no horizontal acceleration. The horizontal component of speed is always 275 m/s and the horizontal distance the bullet moves in t seconds is 275t meters.

There is, vertically, the force of gravity so there is a vertical acceleration of -9.8 m/s^2. The vertical speed, after t seconds, is -9.8t (there is no initial vertical speed). So what will be the height of the bullet after t seconds? What is t when the bomb hits the ground (height= 0)?

"a) How far from the point vertically under the point of release does a bomb hit the ground?"
Is that really what the problem says?? The airplane was at 3000m so the ground is 3000m below the point of release! I suspect the problem really says "horizontally", not "vertically". Find the time until the height of the bomb is 0, and put that into the horiontal distance formula to find the horizontal distance from the point of release to the point of the bomb.

"(b) If the plane maintains its original course and speed, where is it when the bomb hits the ground?"
You are told that the plane is flying at 275 m/s so the distance it flies in t seconds is 275t meters. Use the time until the bomb hits the ground to find where the plane is. You should not be surprized at the answer.

"(c) 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 t he sight at the time of release?"
You have found the horizontal distance the bomb moved and know the vertical distance is 5000 m. $tan(\theta)= horizontal/vertical$

 

[Nickie]

I understand your frustration when faced with a difficult problem. Solving problems in science often requires a combination of knowledge and critical thinking skills. In this case, the problem involves relative velocity and projectile motion, so it is important to have a good understanding of these concepts before attempting to solve it.

To solve part (a) of the problem, you will need to use the formula for the horizontal distance traveled by a projectile, which is given by d = v*t, where d is the distance, v is the initial velocity, and t is the time. In this case, the initial velocity of the bomb is the same as the horizontal velocity of the bomber, which is 275 m/s. The time it takes for the bomb to hit the ground can be found by using the formula t = sqrt(2h/g), where h is the initial height and g is the acceleration due to gravity (9.8 m/s^2). Once you have calculated the time, you can plug it into the first formula to find the horizontal distance traveled by the bomb.

For part (b) of the problem, you will need to use the same formula for horizontal distance, but this time you will need to calculate the time it takes for the bomb to hit the ground relative to the position of the bomber. This can be done by dividing the initial altitude of the bomber (3000 m) by its horizontal velocity (275 m/s). This will give you the time it takes for the bomber to reach the point vertically under the point of release. You can then use this time to calculate the horizontal distance traveled by the bomber, which will give you its position when the bomb hits the ground.

Part (c) of the problem involves finding the angle at which the bomb sight must be set in order to hit the target seen in the sight at the time of release. This can be solved by using trigonometry and the concept of relative velocity. The angle can be found by using the formula tan(theta) = vrel/vel, where vrel is the relative velocity between the bomber and the target, and vel is the velocity of the bomb. The relative velocity can be found by subtracting the horizontal component of the bomber's velocity (275 m/s) from the horizontal component of the target's velocity (0 m/s). Once you have calculated the angle, you can adjust the bomb sight accordingly.

In summary, to solve this problem, you will need to use a