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Question 40.8 ptprojectile that is fired with an initial velocity of 132 m/s is inclined upward a an angle of 519. It lands ata point -28 m lower than the initial p...

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Question 40.8 ptprojectile that is fired with an initial velocity of 132 m/s is inclined upward a an angle of 519. It lands ata point -28 m lower than the initial point. Calculate the time of flight

Question 4 0.8 pt projectile that is fired with an initial velocity of 132 m/s is inclined upward a an angle of 519. It lands ata point -28 m lower than the initial point. Calculate the time of flight



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(II) A projectile is shot from the edge of a cliff 125 $\mathrm{m}$ above ground level with an initial speed of 65.0 $\mathrm{m} / \mathrm{s}$ at an angle of $37.0^{\circ}$ with the horizontal, as shown in Fig. $3-35$ . (a) Determine the time taken by the projectile to hit point P at ground level. ( $b )$ Determine the range $X$ of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point $P$ , find $(c)$ the horizontal and the vertical components of its velocity, $(d)$ the magnitude of the velocity, and $(e)$ the angle made by the velocity vector with the horizontal. $(f)$ Find the maximum height above the cliff top reached by the projectile.

Okay, we have a projectile fired at a 30° angle at a speed of 520. Now we're asked quite a few questions about that scenario, but first let's remember some of our formulas. So remember your co sign component is going to be in your eye direction for velocity and your sign component will be in your J. You also have gravity. So as you have a positive velocity going up, gravity is pulling you down. Now for position you consider that those velocities need to be multiplied by time. You've also integrated your velocity, so you end up with a one half G T squared and then if you also have a initial position, you would also have that in the in the vector. Now there's also equation for range, it's visa, not squared times assign a tooth data all over G. So our first question asked us about range. So fortunately we do have this range formula that we can use and so we'll be putting in our 520 and squaring it And multiplying by the sign of two times are 30° angle, all divided by gravity of 9.8. So when we multiply that out we get 23,895 and that would be meters. Okay? So now the next question is about the time of flight. Well so this range is also um you know that was if we didn't have t we use that value but we have another equation that tells us how far we've moved in the I. Direction and thats rvs of not cosign theta times T. So we can use our 5 20 times the co sign of 30 times T. Will equal that 23,895. This allows us to solve I. T. By dividing both sides by that 520 co sign of 30. So we end up with a range time of five or 53.06 seconds. So if you consider that, that's how long it took for the whole time of flight. If we're going to find the greatest height reached, um we do have a symmetric parabola and so at half the time we can find our height. So if we take that 63.06 and divided by two we can put that time into our position vector. And when we do that you can see that. We also have to then subtract our acceleration component times at times squared and we get a maximum height of 3449 m.

Isn't this equation. We can see that we have negative 16 T square. Our initial velocity is 72 feet per second our initial high zero because it's from ground level. So I guess you could say plus zero. And we want to know when it's at least or greater than or equal to, Ah, 37 feet above the ground. Okay, so we could just, uh, look at the equation 16 t square minus 72 t plus 37. And we're looking for one that is less than or equal to zero because it divided by a negative one to kind of make my leading coefficient positive graphics and a calculator just to see what this looks like. 16 x squared, minus 72 acts plus 37. Okay, so it's going to be crossing the X axis at 0.592 That means our T values gonna range from 0.592 be less than or equal to t, which is less than or equal Teoh. And then it comes back down again at 3.90 seconds. I'm just looking for where it's crossing the X axis. Thank you very much.

You threw an object with 1 41 meter per second square. I'm sorry. Meter per second vertically. It was simply go up and then all back down. No, what would be its velocity? At the highest point of its tragic Terry, it was simply be zero because its initial X component of velocity zero and it's why competent of Alaska become zero at the highest point. So the velocity of simply zero Now what happens if it were? If it went for Fred agrees instead. Now, one thing to note here is that route to is one point for one, which is why you you have the number 1.1 40 meters per second here. Now, if the total of lost total speed is 1 41 meter per second, the hard central component will be 1 41 cause 45 it's will be 100 meters per second on the vertical conference will also be 100 meter per second. So obviously at the highest point, the vertical component with zero on all of its velocity will be the cars under confident and the hearts under competent would not have changed because clarity does not affect it. It means that the highest point, the velocity will simply be 100 meters per second

So for these problems, we know our initial velocity, and we know that an object is fired vertically upward and we have its initial position. So we want to find the position and velocity functions for all times. Um, So what we do is we have our velocity function, and then if you take the, um, if we have an acceleration function, that is, that's what we're going to start off with. So we have our acceleration function with respect to time. When we take the integral of that with respect to t, we end up getting our velocity plus c Well, here's the issue. This plus C is causing issues. So in order to get rid of plus C, we need to use our initial condition. The fact that V of zero equals V not means that now we have is, um, RV of T is equal to, um RVF T is equal to this. Right Here are acceleration function plus c. Well, when we plug in zero where we end up, getting is just V, not Equalling. See? So since C is RV not value now, we have v f t equals our function plus v Not then we can take the integral this with respect to t. And then what we end up getting as a result is our position function. So it's S f t. And then again we want to use our initial condition because it's gonna be plus C so we would have our function and then plus r s not and remember that in this case, Vienna is 29 4. Esna is 30 m, Um and we use all this information to find the highest point. If we look at the position function, we look at the trajectory, we see that right there is the highest point. Um and then we can determine that based on the fact that the slope of the tangent line is zero. So if we set velocity equal to zero software T, we get our value. And then at that time we plug that back into our velocity equation where v f t equals zero and we end up getting that height right here


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