In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 m from this point (Fig. E3.19). If you toss the coin with a velocity of 6.4 m/s at an angle of 60° above the horizontal, the coin will land in the dish. Ignore air resistance. (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?

Question

A person tossing a quarter at 60° to hit a dish 2.1 m away, aiming to win a stuffed giraffe.

Accepted Answer

Hey everyone today, we're dealing with the problem about projectile motion. So we're being told that at a specific arcade, when you throw a ball that is 1.5 m away from you and let me highlight that when it is 1.5 m away from you, throwing the ball at a velocity of 30 m per second at an angle of 25 degrees above the horizontal, the ball will go through the ring. So based on this information, we're being asked to find the vertical component of the velocity of the ball just before it passes through the ring. So in other words, let's think about this in the arcade game and let's draw this out too. So this is the starting point. And let us say this is worthy ring is the sweater. The ring is so this distance here Is 1.5 m. So 1.5 m, let's write that out as. So the change in the horizontal distance is 15 m and I wrote it like this because remember we're dealing with projectile motion. If we're launching an object, it will follow a parabolic path and eventually land at its designated landing area. However, because it's being launched at an angle at an angle theta above the horizontal, which in this case is our X. And this is our projectile. We also have to factor in that there is indeed going to be a vertical component to its path. There's a y. The vertical aspect. Now we're only given the initial velocity of the of the ball as well as the angler which is thrown and the distance that the target is from it, or rather the distance that it travels. So let's redraw this real quick. Just so we can have a better understanding as well because we're given the initial velocity. However, the parameters that were given here, the angle from the horizontal will only help us find a few things because when calculating the horizontal component of velocity and projectile motion, there's only one equation that we can use and that is that the change in distance, the displacement is equal to the initial velocity in the X direction, multiplied by the time of the travel. Now the velocity in the X direction is very simple velocity in the X direction is constant is constant. But so this means that the initial velocity will be the same throughout. It will always be that velocity from start to finish. However, to get there, we have the component velocity, the component initial velocity of 30 m per second. To convert this to the velocity in the X direction, we need to take the initial velocity, the component velocity and multiply it by Kassian data. And the reason for this is that if we are to take the projectile path, imagine it sort of as a right triangle up to a certain point. Then if this is the angle data and this is the hypotenuse then the horizontal which is here, this is the horizontal lies adjacent to the angle it lies adjacent to the angle. So if we read at our trig identities, we have so to when the side in question is lying adjacent to the angle we use Kassian to help us solve for it. So that's why we use co sign data here. So solving that out, we get 30 m per second meters per second, multiplied by Kassian 25°. Kassian 25° which gives us A value of 27.19 m/s. Substituting this back into our equation for all our known values, we have the Distance traveled 1.5 m The initial velocity X. direction, just 27.19 m/s. Multiplied by time. So therefore time is equal to 1.5 m 1.5 m divided by 27.19 m/s, blaming us with an answer of 0. seconds. So this is the time taken to travel 1.56 m in the horizontal direction. But we're not done. We want to find the vertical component of the velocity and to do this, we now need to utilize the values we have and a couple more formulas to go ahead and figure this out. So now considering the vertical aspect, because we do need the initial velocity in the horizontal direction. Excuse me, It is opposite to the angle theta and when its opposite we can use sign to help us find it. So similar to how we did it for the velocity X direction. We can find the velocity in the Y direction. The initial velocity in the y direction by multiplying the component initial velocity and multiplying it by sign data Which equals m/s. Multiplied by sign 25° which gives us a value of 12 0. meters per second. The final equation that we need to use now is that the final velocity or sorry? The final velocity, what we're solving for is equal to the initial velocity in the vertical direction. Of course, because now we're finding the vertical component of velocity before it passes through the ring. So the final vertical velocity plus the acceleration multiplied by the time of travel. Now recall in the vertical direction here in the horizontal direction at least acceleration is equal to zero because the velocity is constant. However, velocity is changing as a projectile reaches its highest point it goes up and it comes down and at all points it is acting against gravity either against or with gravity. So the acceleration experience in the vertical direction, acceleration in the vertical direction, Z. Colton negative G. Stickleton negative 9.8 m/s squared. So with all this information we can go ahead and solve for the final velocity in the vertical direction is equal to and I'm gonna leave at our uh units. For the sake of brevity this will equal 12.68 m/s plus -9. Multiplied by 0. seconds and simplifying that we get that the final velocity in the Y direction Is 12.13 m/s. Therefore, the vertical component of the velocity of the ball just before it passes through the ring is option B 12. m/s. I hope this helps. And I look forward to seeing you all in the next one.

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