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A rifle is aimed horizontally at target 45.0 m away: The bullet hits the target 1,90 cm below the aim point:Part AWhat was the bullet's Ilight time?You may wan...

Question

A rifle is aimed horizontally at target 45.0 m away: The bullet hits the target 1,90 cm below the aim point:Part AWhat was the bullet's Ilight time?You may want to review (Pages 86 90)Express your answer with the appropriate units_0.07SubmitPrevious Answers Request AnswerIncorrect; Try Again; attempts remainingPart BWhat was the bullet's speed as it left the barrel?Express your answer with the appropriate units_

A rifle is aimed horizontally at target 45.0 m away: The bullet hits the target 1,90 cm below the aim point: Part A What was the bullet's Ilight time? You may want to review (Pages 86 90) Express your answer with the appropriate units_ 0.07 Submit Previous Answers Request Answer Incorrect; Try Again; attempts remaining Part B What was the bullet's speed as it left the barrel? Express your answer with the appropriate units_



Answers

A rifle is aimed horizontally at a target 50 m away. The bullet hits the target $2.0 \mathrm{cm}$ below the aim point.
a. What was the bullet's flight time? b. What was the bullet's speed as it left the barrel?

Projectiles fired in the target. It drops it aimed. It's aimed right at the center of the target. That's 15 m away. If you're given the 50 m and we're told it strikes the target actually two centimeters below where where it was aiming, I would convert this quickly. 2 m. So 0.2 meters divided by 100 on DSO. That gives you the Y value for how far fell vertically and the 50 m is the X value in part A. They asked for the bullet flight time, so T equals question mark. I would solve this using the vertical part of the problem because we know several things. You know the vertically the acceleration is due to gravity's or negative 9.8 meters per second. Squared the initial vertical velocity or call it V I. Why is zero Because it's fired horizontally at the beginning and the displacement? Why is negative point negative 0.2 m Negative Because it's going down, I should be able to solve for the time. Then, using this information, I'm gonna use a kingdom attics equation, the one without, uh, the final in it. So X equals V I t plus one half a T square. Of course, we're doing vertical stuff. So the ex turns into a why v. I is zero. So you plug that in here zero times. Anything is zero. So that goes away. So you're left with why equals one half a t squared? I plug in what I know. Negative 0.2 equals one half times negative. 9.8 times T squared. Continuing the solution up here negative 0.2 equals negative. 4.9 t squared. We're gonna divide both sides by negative 4.9 to make that cancel out on the right hand side. So this cancels when you do that division, you get positive. Small number 0.0 four equals t squared, and then you take the square root of that and you get tea is equal to 0.6 seconds. Yeah, so that should be the time it takes for the project. How to reach the target because that's how long it was actually dropping for part B. They ask, what's the bullet speed when it leaves the barrel? So for this, we're gonna look at the horizontal part of the problem and we use the equation. V equals X over tea. We know access 50 m divided by 0.6 seconds. You divide that out and you get okay. 833. I'll just rounded off 830 meters per seconds. That that should be the speed with which it leaves the gun. That, of course, would be v sub x speed with which it leaves.

So you know, the Y coordinate of the bullet is equaling negative 1/2 gt squared and this is because the initial wild velocities equaling zero. And so we can say that if why is equaling negative point 019 meters for part A. Ah, we can say that T is equal in the square root of negative too. Why over g? So this would be equaling a square root of negative two multiplied by negative 20.19 meters and then this would be divided by 9.8 meters per second. Squared and T is than Equalling 6.2 times 10 to the negative second seconds. So this would be our answer for part A. For part B. The muzzle velocity is the same as the initial horizontal velocity of the bullet. Because exes equaling 30 meters, we can actually say that the exes equaling 30 meters and we know that this is equal in the initial velocity multiplied by tea. So to find the initial velocity, which is again all in the ex direction, this would be equaling X over tea. So we have 30 meters and then this would be divided by 6.2 times 10 to the negative second seconds and we find that the initial velocity is gonna be equaling two, approximately 480 meters per second. This would be our final answer for part B. That is the end of the solution. Thank you for watching.

All right, I went through this but I made a mistake. V equals V. zero. Which is this? I shouldn't be multiplying by time. V equals v. zero plus a. T. So Um v equals v. zero plus a. T. And it's not one half a. T. It's just a T. Okay, that gives me two V. Signed data over A G. Is T. Okay, now, so that's what I got from above. And I'm using the fact that At the peak the velocity is zero and at the peak time is half of the total time. Okay, now, below here, X equals V. Initial in the X. Direction times. T. Just solving that, I get T. Equals X over V. Cosign theta. And I put it up here. So now I can solve these two that will give this and this and so it gives this and that gives um putting that on a calculator. Try to find my calculator. Okay, It's gonna be one af inverse signed of X one G over V Squared uh zero zero point 606 degrees. But that's not the answer because we want to know the height above the target to aim. So that would be the tangent. Um So d is going to be uh X. Tangent data, which is not this it's going to be 4.8 four centimeters. Thank you for watching. Mhm. Mhm.

A bullet is shot horizontally from shoulder to shoulder height, which is 1.5 m within initial speed of 200 meters per second. How much time elapses before the bullet hits the ground? How far does the bullet travel horizontally? So if we're talking about the change and why? So remember the change and why is the velocity in the Y direction plus one half GT squared? There's no initial velocity in the Y direction, So this gives us T. is equal to the square root of two times 1.5 Over 9.8, which is zero 55 seconds. And in part B, how far does it travel horizontally? Well horizontally? It travels VC. Got X times T. So 200 m/s times .55 seconds Is 110 meters.


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