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