A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in FIGURE EX4.13. The ball's position is shown at 1 s intervals until t = 3s. At t = 1s, the ball's velocity is v = (2.0 i + 2.0 j) m/s. (a) Determine the ball's velocity at t = 0 s, 2s, and 3s.

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Hey, everyone in this problem, we have astronauts on a space mission. Hey, they're gonna throw a projectile as part of this mission on an unknown planet that follows the projectile path shown below. So we're given this diagram, we have the X position in meters on the X axis, the y position in meters on the Y axis. We kind of have this parabolic type shape. We're shown that at two seconds, the ball has a velocity of three I plus five J and then we're shown intervals 02 seconds, four seconds and six seconds. And we're asked to calculate the velocity of the projectile at zero seconds, four seconds and six. We're given four answer choices. Each of them having a different velocity for the V zero V four and V six. OK. And meters per second. And we're gonna come back to these answer choices as we work through the problem. Well, let's start with what we do know. Mm Well, let's think about the horizontal direction, right? And we're gonna assume that the horizontal acceleration is zero. Hey, there's no air resistance or anything like that acting, which is a typical assumption. So we're gonna assume that the horizontal acceleration is equal to zero. And if we do that, that tells us that the X component of the velocity is constant, we know that the X component of the velocity at two seconds is three. OK. And so the X component VX is going to be three meters per second for all time points. All right. So we're done with the X component. Now, we need to worry about the Y component. And if we look at our answer choices now that we've done the X component, you can see that that doesn't narrow it down every single answer choice. Every single time point has three I meters per second, that horizontal component of the velocity. And so this was the correct assumption to make and now we can move forward. Yeah, what other time point can we use to figure out more information we're on an unknown planet. So we don't know the acceleration in the vertical direction. So if we're trying to use our kinematic equations, we don't really have enough information. What we do know is that at four seconds, this is the maximum height, they were at that maximum height and that tells us something about the velocity. OK. So let's look at the time period from two seconds to four seconds, see if we can figure anything out there. So from two seconds, so four seconds, OK, the initial velocity in the Y direction. Well, that's gonna be the Y velocity at two seconds, which we're told is 5 m per second. The final velocity in the y direction is going to be 0 m per second. Hm. Because we're at that maximum, remember that when a projectile is at its maximum height, the velocity is going to be zero momentarily as it switches directions, right? So we have that final velocity of zero. We don't know the acceleration because we're on an unknown planet. And we know that the time taken is two seconds. I had to go from two seconds to four seconds. That's a two second interval. So now we have three known values, we wanna find the acceleration so that we can use it for the other intervals. We can do this. OK. So we can choose the kinematic equation with those four variables. And that's gonna be VF is equal to V na plus 80. Substituting in our values, 0 m per second is equal to 5 m per second plus a multiplied by two seconds. So we get that negative a multiplied by two seconds is equal to 5 m per second, which tells us that our acceleration is negative 2.5 m per second squared, they dividing by two seconds on each side. So now we know the acceleration due to gravity on this unknown planet or the acceleration that the projectiles undergoing in the y direction. We're gonna put a red box around it negative 2.5 m per second. Squared, right. So that's helpful because now we could switch over to our other intervals. So let's move to our next interval and we need to figure out the speed at zero seconds. Ok. So let's look at the interval from 0 to 2 seconds. Again, we have information at two seconds. So we wanna compare everything to that two second time. Hm. So from zero seconds to two seconds and actually let me hold off for a second before we do that. Let's go back to this four seconds because we can write that velocity out. We had that the X component is always gonna be three. And then we said that the Y component is going to be zero there because it's momentarily at rest as it switches directions at its maximum. OK. And so the velocity at four seconds is going to be three I plus zero J meters per second. And that is the given velocity for all answer choices. OK. So that doesn't narrow down our answer at all. Um But that is the correct velocity that we found that matches the answer choices. So we're good to go to find those other two velocities at zero in six seconds. All right. So back to our 0 to 2 2nd high point. Now, the initial velocity V not why this is going to be that velocity we're looking for, this is the Y component of the velocity at zero seconds. The final velocity is gonna be the velocity at two seconds in the Y direction which is 5 m per second. They don't mix up our initial knot with the zero. OK. We now know the acceleration negative 2.5 m per second squared. And we know that this takes two seconds. OK? We're going from zero seconds to two seconds. That's an interval of two seconds. So again, we have the same four variables. We're gonna use the same equation via is equal to V not plus 80. Substituting in our values 5 m per second is equal to V not plus negative 2.5 m per second squared, multiplied by two seconds. That tells us that 5 m per second is equal to V knot minus 5 m per second. And if we add 5 m per second to both sides, we get that V knot is going to be 10 m per second. OK? So now we can write our speed at zero seconds. V zero. This is gonna be three, I, OK. The same horizontal component and we have a vertical component of positive 10. So three I plus 10 J. OK. Let's take a look at our answer choices now. And what we can see is that we can narrow it down to B already. Option A has three, I minus 10 J. OK. So the wrong sign option C and D both have 7.5 as the magnitude of the Y direction which is not correct. Right. So it looks like me is gonna be the correct answer. But let's go ahead and find the speed at six seconds as well. Just to be sure we haven't made a mistake to see how to do that process as well. Right. Let's keep going. So now we're gonna go from, let's say, from four seconds to six seconds and there's a couple different ways you could do this. We have the velocity at zero seconds and at two seconds and at four seconds now. So you could use any of those. If we look at our diagram, we kind of expect the speed at six seconds or the velocity to be the same, but in the opposite direction. OK. So we expect the magnitude to be the same but the sign to be different. Let's see if that's true. So our initial velocity in this case is gonna be that velocity at four seconds, which we know is 0 m per second in the Y direction. The final velocity is what we're looking for. The acceleration again, negative 2.5 m per second squared. And the time is again a two second interval using that same equation that we've been using VF equals V, not plus A, we have VF is equal to 0 m per second plus negative 2.5 m per second squared multiplied by two seconds. And we get the final velocity is negative 5 m per second in the Y direction. Which again, oops, let's not do that yet, which again is the negative of what we expected. So this is all great. And so our speed at six seconds, we have three I, OK. That magnitude of three in the horizontal um direction that's constant. And then we have minus five J uh minus five J the vertical component we just found. So let's go back to our answer choices and see what we found. Ok. Now, for the speed at zero seconds, we have three I plus 10 J, we think that's gonna be option B the speed at four seconds. We found three I plus zero J, which is the same for all answer choices. And for the speed at six seconds, we found three I minus five J meters per second. And so this does indeed correspond with answer choice speed. Thanks everyone for watching. I hope this video helped see you in the next one.

A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in FIGURE EX4.13. The ball's position is shown at 1 s intervals until t = 3s. At t = 1s, the ball's velocity is v = (2.0 i + 2.0 j) m/s. (a) Determine the ball's velocity at t = 0 s, 2s, and 3s.

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Key Concepts

Projectile Motion

Velocity Vector

Kinematic Equations