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Ch 03: Motion in Two or Three Dimensions

Chapter 3, Problem 3

A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance. (d) How far has the football traveled horizontally during this time?

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Hey everybody. So today we're dealing with the problem about projectile motion. We're told that an angry woman after a fight with her boyfriend tossed a pendant with a horizontal velocity and the initial horizontal velocity of 80 m per second, initial horizontal velocity of 80 m per second and initial vertical velocity of m per second. Using this, we're being asked to calculate the distance that the pendant traveled after it was thrown and air resistance is negligible. So let's break this down a bit before doing any of the math. Let's think about this. Conceptually, let's say she throws it and we have a vertical and horizontal component. And theoretically when we're dealing with projectile motion, we assume that it takes a parabolic path takes a parabolic path, it's a horrible parabola. But anyways, let's say this point right here is max height, it is h max and we have a horizontal velocity which are put in green, a horizontal velocity Horizontal velocity of 80 m/s and a vertical velocity which I'll put in red initial vertical velocity of 50 m/s. So with this in mind, we can also use some principles, some facts about projectile motion and that is that the acceleration in the wider or in the X direction is zero because because velocity vernon X, he's constant. So if velocity hasn't changed and that means there is no acceleration, however, in the y direction it's acting against gravity. So the at least all the way up until the h max. So until h max until here the Acceleration is the negative force of gravity which is negative 9.8 m/s square. So with that under our belt we know we can use some trigger metric functions because we have the initial height in the vertical direction. Or sorry, initial excuse me. The initial velocity in the vertical direction. We have the acceleration in the vertical direction and we know that at the maximum height at maximum height right here, if there's a ball here, let's say the pendant is being thrown at maximum height is when it stops its vertical ascent and begins its vertical descent. So at that point since direction is changing, it halts in mid air for a moment. So that means that the velocity here And we'll just say final because it'll make sense in a second when we do our calculations the final velocity at least at maximum height is 0m/s. So we have these three values and we want to find the time. Why? Because to find vertical displacement. Or sorry, horizontal displacement, which is what we're looking for. We're looking for the distance that dependent travels. We need two functions. We already have one of them. It's vena X. But we need time and time can't be solved in the in the horizontal direction without the vertical component. So let's use a uniformly accelerated motion equation, one of our kingdom attics equations to go ahead and solve for vertical or use the vertical component to solve for time. So let's use the function that the final velocity is equal to the initial velocity plus A. In the vertical direction. Multiplied by T and T is what we're looking for. So substituting in our values we have zero is equal to 15 m/s Plus or sorry minus plus, sorry plus negative 9.8 m per second squared in two time. Let's write that in black Solving four time. We therefore get That time is equal to 50 m/s divided by 9.8 m/s sq Which gives us a value of 5.10 seconds. However, we're not done recall that this is only the time it takes. Alright it in green. This is only the time it takes to reach maximum height To reach the end point, it has to be two times the time it takes to T. Is therefore equal to 10.20 seconds. And now that we have the actual T. Which is t total the time it takes from the pendant being thrown to the time it reaches the ground we can go ahead and bring that over here. So let's sub in our values delta X. Is equal to the initial velocity in the X direction which is 80 m/s, multiplied by Time, which is 10.20 seconds. Which gives us a final final answer of 816 m 16 m. Therefore the distance that dependent that the angry woman through Is indeed answer choice c. m. I hope this helps, and I look forward to seeing you all in the next one.
Related Practice
Textbook Question
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?

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Textbook Question
A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance. (b) How high is this point?
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Textbook Question
A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance. (c) How much time (after it is thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)?
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The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (See Nature, Vol. 424, July 31, 2003, p. 509.) (a) What was the takeoff speed for such a leap?
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On level ground a shell is fired with an initial velocity of 40.0 m/s at 60.0° above the horizontal and feels no appreciable air resistance. (b) How long does it take the shell to reach its highest point?
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