Projectiles Problem - hardest physics problem I have ever faced

[lolzwhut?]

Ok, here are the last 2 problems of my practice packet. I honestly, have no clue on the face of Earth how to even get started. If someone can guide me through a procedure. Tell me what to do, I can complete it myself. Honestly right now, i have no clue how to even start this problem. There are 2 parts to it!

Homework Statement

A bomber flies horizontally with a speed of 284 m/s relative to the ground. The altitude of the bomber is 1470 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.


Horizontal Speed: 284 m/s
Altitude: 1470 m
Acceleration of g: 9.8 m/s^2

Homework Equations

NO CLUE, I AM REALLY CONFUSED!

The Attempt at a Solution

I AM CONFUSED, what should I start this problem off by?

[Part 2]

Homework Statement

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 degrees.

Homework Equations

I believe in this part I have to take the arc tangent. But I'm not sure of what because I do not know how to do the first part. Please help!

The Attempt at a Solution

Need help on the first part to solve this part.

 

[ApexOfDE]

I think problem asks you to find the distance in x-axis between the initial point (when bomber is realsed) and the point when it hit the ground. Because you have speed in x-axis, you just need time t to find distance. When released, vertical speed is zero (v0y = 0) and you have altitude (h = 1470m), now you can try to find vertically termin speed => t.

p/s: i am thinking about p2...

 

[lolzwhut?]

I think problem asks you to find the distance in x-axis between the initial point (when bomber is realsed) and the point when it hit the ground. Because you have speed in x-axis, you just need time t to find distance. When released, vertical speed is zero (v0y = 0) and you have altitude (h = 1470m), now you can try to find vertically termin speed => t.

p/s: i am thinking about p2...

I'm still not clear about the concept :S

So since it wants us to the find the Yfinal after finding time and distance?

 

[ApexOfDE]

No, it asks you find the distance between 2 points in x-axis (horizontal). The former is the point the bomb released (x = x0, y = 1470m) and the latter is the point the bomb hit the ground (x = x(t), y = 0).

 

[lolzwhut?]

What formula am i supposed to use for that?

Displacement formula?

 

[ApexOfDE]

can you calculate vy(t) if you have vy0, h and g?

 

[lolzwhut?]

can you calculate vy(t) if you have vy0, h and g?

need acceleration, and time?

 

[BlueEight]

When the bomb is dropped, it will have a velocity of 284 m/s, which will be SOLELY horizontal. Because of independent superposition, the fall of the bomb will not be affected by its horizontal movement. Therefore, solve for the time that the bomb will take to reach the ground as if it were in free fall, and multiply the time by its horizontal velocity to obtain the horizontal distance traveled.

 

[ApexOfDE]

You know this formula: v(t)^2 - v0^2 = 2as, don't you? Use it to find v(t) and remember that this v is a vertical speed.