a-ball-is-thrown-in-a-projectile-motion-trajectory-with-an-initial-velocity-v-at-an-angle-theta-abov

Question

A ball is thrown in a projectile motion trajectory with an initial velocity $v$ at an angle $	heta$ above the ground. If the acceleration due to gravity is $-g,$ which of the following is the correct expression of the time it takes for the ball to reach its highest point, $y,$ from the ground?
(A) $v^{2} \sin t / \mathrm{g}$
(B) $-v \cos 	heta / g$
(C) $v \sin 	heta / g$
(D) $v^{2} \cos 	heta / g$

Sachin Rao · Accepted Answer

this problem, we have given a ball is thrown in a projectile motion trajectory with an initial speed. We at an angle theater of the ground. This has gone the same well is trita have the ground. If the excavation due to gravity is minus G. We know that in the downward direction. Which of the following is the correct expiration of the time it takes for the ball to reach its highest point. Right? So we know that this is supposed highest point. Right? So we have to find this time. So we know that total time. We can find from this point from point A. To point B. We can find the time and this time will be equal to the half of that time was light. So the total time was light. We can find it is a call to We know that as y is equal to you White B. Let&#x27;s have where it is where we will use this kind of attack occasion. So first we will find that total time of light says when violence land on the ground, The vertical displacement. Will we call to zero from point a to week vertical displacement 20. And initially speedy&#x27;s recent data into feet and this is half into G. P square. So we can find the their time. It is a cool do he is a call to to be signed by G. This is the total time of slide. Great. No. We have to find the time from the initial point to the highest point. So this will be equal to the half of this time. So our time strong initial point to the maximum point where we called T baidu or we can say the scientific obliging. Now we will check the options. Option C. Is correct. We can write option C. Is correct.

Ze-Han Lee · Answer

right. So in this problem, we&#x27;re throwing a ball. Was initial velocity V and an angle from the horizontal is data. So we want another time that from the initial position to the highest point. So want Thio. Find out the time from this position to this position. So the highest point means that Ah ah, So it is the location. Weird. Why the velocity along the wind direction zero. Right. So we can simply utilize v times Cynthia which gives the initial velocity allowing the Y direction in minus zero, which is the velocity at the highest point along the Y direction right? And this equal with g times t. So you can see that the total time is a V sign thehe blog, all right?

Henrique Saito · Answer

So this question involves a projectile that is being launched up at a velocity of V I at an angle theta above the horizontal. And, you know, obviously we have gravity as the excel as the vertical acceleration of negative 9.1 meters per second squared or negative G. And you know there&#x27;s no horizontal acceleration. And so the first thing we gotta do is find out the horizontal and vertical velocity at the peak. So this is the peak. So the horizontal velocity, What&#x27;s that gonna be? Well, since there&#x27;s no horizontal acceleration, the horizontal velocity is going to be equal to the initial horizontal velocity. Right? And so what is that gonna vehicle to? Well, that&#x27;s just gonna be equal to the initial velocity. If you look from the struggle, weaken. See, that&#x27;s going to the initial velocity. Times Co signed data, right? So that&#x27;s that. And now for the vertical velocity. Well, as we know, as soon as the projectile reaches his peak, that is the exact moment when vertical loss equals 20 OK, so that&#x27;s our vertical velocity peak and blues. Their results velocity Next with the find out. When the project, the projectile reaches to speak. Okay, So for use this equation of the equals to, um v not plus a t. Well, if we use this for the y component, we know that V Y is gonna be equal to zero. You know, the vino Y is gonna be equal to, you know, again using this triangle where this is. We know why have you not? Why is going to be equal to V? I signed data. A wise is gonna be with, uh, negative g antes de this is equal to zero. And then we just solved for tea. So here we have t equals V I Science data over G. So this is the amount of time takes to re speak, and that&#x27;s that. And next finally, we have, uh you have to figure out this maximum height H So what is this h going to be? Well, now that we know the time, we can use a second formula where we have Delta X, which is going to be the change in displacement, which is really represents the height. If we&#x27;re looking at this in the y component, Delta X is gonna be equal to view why, plus the uh, not why? Over to times time. So let&#x27;s just do some replacements. We know view is gonna be equal to zero. Okay, You know, we not why is equal to, uh, uh v I science data. Sorry. And we know that t is gonna vehicle to V. I signed data over G. So if we sort of replace those, we get Delta X, which equals the height is gonna be equal to V. I signed data over two times V. I signed data or G. Right, Because that&#x27;s time. And so then we just do some cleaning up. We&#x27;ll get V. I signed data squared over to G, and that&#x27;s your answer.

Petros Markopoulos · Answer

That&#x27;s all this problem about projectile motion from some point that we will call a we launched a projectile an initial velocity B that forms an angle local fada with the ground. The longest parabolic trajectory be is the highest point where the velocity is now. Hugh. What we want to know is what the magnitude of this velocity is. What the speed of the projectile at point B is to do that it will re examine things. Appoint a redraw A with the initial velocity. So these a vector and we can decompose it toe horizontal component in a vertical component V one and V t. And here I need to remind you that the angle here is fatal. Using simple trigonometry, we can calculate these lengths. The length of the one is equal to the length of V times coasting pita because V one forms the adjacent side to the angle Fada in this right triangle. Similarly, the length of the two, the sequel to the length of the times Sign of feta. If you look at the trajectory again, we can now use these to find. The answer here is be And here is his philosophy. You again. If we look at the vertical component V to, it starts pointing upwards, and as we go along the trajectory, you can imagine it growing smaller and smaller, and Philip points down downwards and the projectile starts following. So what? Their point right before that, the highest point in the trajectory. It has to be zero. It has to not exist. On the other hand, the horizontal component has no force acting on it. There is no acceleration or deceleration in the horizontal direction, so via one remains unchanged. And since there&#x27;s no vertical component at you, the only thing that remains is V one. So we know that you was equal to be one, and therefore the magnitude of you is equal to the manager to be one. So they speed. There is fi times coasting Fada that is answer D from what is given to us. And that is the final answer for how we looked at the vertical component going to zero. There is a mathematical explanation through differential calculus. If anyone wants to look into that, it is very interesting, but it&#x27;s not necessary for the S a. T, because this intuition works just fine.

Jayashree Behera · Answer

The horizontal velocity over here remains constant during the flight because there is no acceleration in the horizontal direction. So final velocity along X direction left us colic. V X is going to be holiest initial velocity along the X direction. That&#x27;s the horizontal direction and this is going to be very not cost. Tita Martita is the angle made by the project, with the horizontal now at the top off the flight, the vertical velocity become steel. So if this is the X axis and this is the lie axis and the Project, I it&#x27;s like this. So at this point, top off the flight, the vertical velocity goes to zero. However, that is, some horizontal component of velocity is still in this direction, for which the wall continues to move horizontally. On. That demon&#x27;s unchanged because, as I said, there is no acceleration in the hottest under direction. So the velocity in the horizontal direction women&#x27;s constant constant arguments unchanged, so the correct option is going to be option D

Caitlyn Axe · Answer

this problem were given a height above the ground and an initial velocity of an object that it&#x27;s turned into the air and lands on the ground and were asked to find the speed of the object when it hits the ground in these terms of h E and our angle. So to start, we are going Teoh When the initial energy of the ball, we know that energy can be kinetic energy and gravitational potential energy. The gravitational potential energy of the object on the top of the roof is equal to the mass of the object times. Gravity tons the height above the ground, which is h were given that value in the problem and the kinetic energy of the object is equal to 1/2 turns, the maths times, the initial velocity of the object squared and now we&#x27;re going to find the final energy of the objects. So that is the, um ball. When it hits the ground again, it is kinetic energy plus gravitational potential energy. And we know that the gravitational potential energy is equal to the mass of the object times, gravity times the height of the object, which, when it hits the ground is now zero, sir. Gravitational potential energy of zero. We&#x27;re left with kinetic energy, which is equal to 1/2 mass. The object times are final velocity squared and we&#x27;re gonna set these two equal to each other because we know that energy is conserved throughout, uh, scenario. So we have won&#x27;t have. And the square plus n g h equals 1/2 and B f Square. And we have mass on both sides of this equation, so you can cancel it out and we&#x27;re left with 1/2 of these squared plus g h equals 1/2 pf squared. We can multiply both sides by two and we end up with you squared plus to GH he calls the final square. Well, you will take the square root of both sides and we end up with you. Final equals the square root of our initial velocity squared plus two jeanne each that it&#x27;s the answer to part

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