A shot putter releases the shot some distance above the level gro...

Question

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. (c) How far did she throw the shot horizontally?

Accepted Answer

Hey everyone. So today we're dealing with the problem about projectile motion. So we're told that an athlete who is a few meters above ground level is throwing a javelin at a velocity An initial velocity of 20 m/s and they're launching the javelin at an angle of 46° above the horizontal. Given that the Javelin landed on the ground 4.11 seconds later we're being asked to find how far the Javelin was thrown or rather the distance traveled by the javelin. So with this in mind, let's go ahead and draw this out. If we have projectile motion, if something is being launched at an angle at an angle or angle theta, then there is a horizontal component, right? The distance that it's traveling, but there's also a vertical component that denotes the height at which the project us as it reached or will reach eventually. So in this case we're dealing with the horizontal a component of our projectile motion because we're given the distance that reaches and hits the ground, which means it's the distance that following a parabolic arc, right? It's the distance horizontally from point A to point B. So with that in mind, we need to use the horizontal component to really calculate our uh, the distance traveled. So the general of thumb, when considering the horizontal aspect of projectile motion, we can say that the change in distance, the displacement in this case is equal to the initial velocity, multiplied by time. So we're already given time, we know that time Is equal to, or sorry, time is equal to 4.11 seconds. However, even though we have an initial velocity of 20 m/s here, that's not what we're looking for. This is the velocity of the projectile itself, including both vertical and horizontal components. Now, if we recall when considering horizontal, the horizontal component of projectile motion, we know that the velocity in the horizontal direction is constant, It is constant, which means acceleration is zero. But because it is constant, we know that the initial velocity has to be equal to the initial velocity in the X direction. Like the initial velocity here. I mean the initial velocity in our equation has to be the initial velocity in the X direction. Because that will be a constant. To find this though. We need to take the horizontal or the initial velocity. V naught and multiply it by sine theta or sorry not sign data, co signed data. And the reason for this is because of our tree geometric rules. We have a horizontal and if we consider this like a right triangle, if the angle is here and this is the hypotenuse, then the horizontal aspect lies adjacent. If we recall our rules are trigger metric identities, we have so so since it is adjacent, we use Kassian times data to reach our answer. So we multiply the initial velocity times cosine theta Which is simply 20 m/s, multiplied by Kassian for 46°, Which equals 13. m/s. Subbing this back into our equation up here. Rather let's bring it down. We can therefore say that the distance traveled is equal to 13.89 m/s, Multiplied by 4.11 seconds, 4.11 seconds, which gives us a final answer. A 57.0ops of 57.09 m. Therefore, the distance traveled by the Javelin, assuming that it lands 4.11 seconds later with the given Parameters is answer choice D 57.09 m. 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. (c) How far did she throw the shot horizontally?

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