A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0° above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. (b) What are the components of the shot's velocity at the beginning and at the end of its trajectory?

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Hey everybody. So today we're dealing with projectile motion. So we're being told that a ball is being thrown from the top of the building to a taller buildings a certain distance away. And were given the initial velocity of the ball Being 14.5 m/s, the initial velocity And we're told that the angle at which the ball is being thrown is at 25 or 21.5° above the horizontal. We're also told that it hits the ground 3.8 seconds later. With this, we're being asked to find the horizontal and vertical components of the ball's velocity initially and at the end of the trajectory, once it just before it hits the ground or just as it hits the ground. So let's really quickly, let's draw this out. Let's draw this out conceptually and right, yeah, let's go. So we have a starting at an end point, we know that the hoops, we know that the initial velocity that it has been launched at The initial velocity this way is 14. m/s. However, it's being launched at an angle This angle of theta, which is 21.5° and that means that for the horizontal component, there's also a vertical component that we have to worry about. So we have a vertical and a horizontal component. So let's go ahead and think about that real quick. The horizontal component, which I'll draw in red. The horizontal component has a velocity like this from start to finish and this is because velocity in the horizontal direction of projectile motion and you should all know this velocity in the X direction stays constant. So it is the same as the initial velocity. This also means that there is no acceleration, acceleration is zero because velocity is constant. Using blue to draw the vertical velocity for vertical velocity for the vertical component. Do we have an initial component and we have a final component? Keep in mind that as it's going up As it's going up a is equal to negative 9.8° and as it's coming down, oh sorry as the as the ball is being thrown it's constantly experiencing in gravity or a force of 9.8 m per second squared which is acceleration which is equal to negative G. Etcetera etcetera. And this is because it's acting against gravity. So on the way up it will be a sort of destructive force. It'll be interfering with the velocity up. And that's why it eventually starts to come back down. Which is when the velocity and the acceleration act in the same direction, that's what happens right here at the end. But with all that in mind, we also need to keep in mind that the sort of triangle that we have here when it's at max height at least can be used to also find the values of the X. And Y components. So for the why the initial velocity. Well p the initial velocity multiplied by sine theta. And this is because using our trick geometric rules. So tour if we think about it, the horizontal component lies opposite to the angle. So since it's opposite we use sign like we do here similarly, the X component will be the initial velocity times co sign data. Why? While again using trigger metric identity, it is the horizontal component is adjacent to the angle with it with which the projectile is being launched. So since its adjacent we use co sign. So let's go ahead and really quickly calculate both of these. So this will be 14.5 m/s, Multiplied by sign 21.5° which gives us an answer of excuse me of 5.31 m per second at release. So that is our initial velocity there in the Y direction and the velocity in the X direction, which again is constant, is calculated in a very similar manner. Temps Kassian 21.5 degrees which gives us an answer of 13. m per second. However, we're not done. We still need to find the final velocity in the Y direction and we can solve this really easily because we have all the necessary components we have the time it takes to reach the ground. We have the initial vertical velocity and we have acceleration. So using our kingdom attics equation, we can go ahead. So the final velocity. Final velocity is equal to the initial velocity plus a times T. So substituting in our values, we get 5. m/s plus -9.8 in two 3.8 seconds. Oops! That went off the screen. I'll just leave all four units for now. So simplifying that we get that the final velocity is equal to -31. meters per second. That is our final velocity. So with all of these, that means that the answer choice that we are looking for is answer choice C. The initial or the horizontal velocity is 13.49 m per second. The initial vertical velocity is 5.31 m per second. And the final velocity in the Y direction is negative 31.93 m per second. I hope this helps. And I look forward to seeing you all in the next one.

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0° above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. (b) What are the components of the shot's velocity at the beginning and at the end of its trajectory?

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Key Concepts

Projectile Motion

Velocity Components

Kinematic Equations