On level ground a shell is fired with an initial velocity of 40.0...

Question

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. (a) Find the horizontal and vertical components of the shell’s initial velocity.

Accepted Answer

Hey everyone today, we're dealing with the problem of projectile motion and uniformly accelerated motion. We're being told that a coin that is lying on level ground is fired so that it acquires a velocity of 24 m per second. An initial velocity of 24 m per second directed at an angle of 38 degrees above the horizontal. With this information, we're being asked to find both the horizontal and vertical components horizontal and vertical components of the initial velocity. Considering air resistance to be negligible. So let's think about this conceptually really quickly before doing any math. If we have, let's say this is a coin. If it's being launched at an angle that means that not only does it have a this is the path. By the way, if it's being launched at an angle that means it not only has a horizontal component, but it has a vertical aspect to it as well. It's moving up as well as to the side that's denoted like so and we're being asked to find these components themselves. But how would we do that? Because we're only given the initial velocity of were only given the initial velocity here, not the different composite parts, different components and the angle that has launched that. Well, this is very simple if we think of the projectile motion at least until the maximum height has somewhat of a right triangle, we can go ahead and use our trigger metric identities to go ahead and actually give us this answer. So if this is therefore the hypotenuse if the complete vector of both the horizontal and the vertical components is the hypotenuse, then that means that the vertical component, the vertical component lies opposite. It is across from the angle data and the horizontal component lies adjacent it is right next to. And if we go ahead and write this out again with our trick geometric identities. So when the when the aspect is opposite the angle, then we have to rely on sign, we have to use sign to calculate that. So we would use sign to to to get the vertical aspect similarly, if it is adjacent to the angle, we would use Kassian. So we would use Kassian data to get the horizontal. So with this in mind we can go ahead and bring this to our initial velocities because we're still not done yet. Let's start with the vertical component, forced forced vertical component. First, the initial velocity in the y direction will simply be the initial velocity given of everything. Multiplied by sine theta, Which is 24 m/s, multiplied by sign Which gives us an answer of 18.9 m/s. Similarly for the X, the horizontal component of the initial velocity we take the initial velocity multiplied by cosine data as we mentioned earlier. So this is 24 m per second, multiplied by Kassian Kassian 38°, Which gives us an answer of 14.8 meters per second. So looking back at our answer choices then if these are our values, We have 18.9 for why here, and 14.8 for the horizontal here. So our answer is a. The horizontal component is 14.8 m/s, and the vertical component of the initial velocity is 18.9 m/s. I hope this helps, and I look forward to seeing you all in the next one.

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