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. (d) How far from its firing point does the shell land?

Accepted Answer

Hey everyone today, we're dealing with the problem about projectile motion. We're being told that a pot on the ground can explosively fire a seat at an initial velocity, initial velocity of 33m/s and it can be fired at an angle of 37° above the horizontal. With this information, we're being asked to find where the seed lands relative to the position of the pod, assuming the distance or assuming that air resistance is negligible. What this means is actually we're looking for what is the total distance that it takes? Total distance it takes for the seed to fly through the air from the pot and land on the ground again. So with this in mind, let's go ahead and sort of draw out a little schematic for him. So we know that projectile motion, it's being launched at an angle an angle theta. So if it's being launched, hey, and the steep path that includes a horizontal component but a vertical component again. Because again it is being launched at an angle, so it is being launched vertically as well as horizontally vertically as well as horizontally. So with that in mind, even though I've drawn this rectangle here, it's important to remember that in projectile motion it follows a parabolic arc. So it's a little inaccurate here, but this is maximum height and this is where it lands again. So we need to find the horizontal distance and to find horizontal distance, we have a singular equation to do. So the change in X. The change in distance, the displacement horizontally, is it called city velocity in the X direction, multiplied by time. Keep in mind that we don't have acceleration. We don't have acceleration in the horizontal direction because the velocity is constant because the X. Is simply the initial velocity of X, which is constant. So with this we can go ahead and determine what the horizontal velocity is because the velocity we're concerned about horizontally requires a bit more math. So looking back at our triangle that I drew here representing sort of the path to the highest point. The horizontal component lies adjacent to the angle. And remembering our trigger metric identities, we have so Cotto. So so if you're dealing with a side that is adjacent to the angle, we have to use cosine to find its angle. So therefore this means that the again because the initial velocity is nothing but the horizontal velocity in this case V. X. Is equal to the initial velocity multiplied by Kassian data. Kassian data which is 33 m/s, Multiplied by Kassian 37 degrees, which gives us a final value of 26.4 m per second. However, we're not done here because we have yes, we have the velocity in the horizontal direction but we still don't have time. So to solve this, we need to actually use the vertical component as well because the uniformly excited emotion equations for vertical motion have time as a variable. So let us figure that out. The way we do this though is a sort of convoluted way. So let's write out the variables that we know. The initial velocity in the y direction is what we need to find. It's pretty simple. We use a similar concept as we did for the vertical or horizontal velocity. The vertical velocity lies opposite the angle, so therefore if it lies opposite, we use sign because opposite there. So this is the initial velocity multiplied by sine theta, which is equal to m/s Time sign. 37, which is equal to 19. m/s. We have our acceleration. We are, it is being launched vertically, which means it is accelerating against the force of gravity. So the acceleration is therefore the negative force of gravity. So it's negative 9.8 m per second squared our final velocity We know, and we're just concerned with maximum height at the moment in time. And this will make sense in a little bit. But at maximum height, at maximum height, the final velocity will be zero. It'll be zero m for a second. And this is because at maximum height as I've drawn here, that's the point where the projectile has reached the top most point. So it has no more vertical velocity, right? It has no more vertical velocity in the up direction, but it hasn't started its descent downward. So it also has no velocity in the down restriction. So it halts in midair just for a second. But at that one point at the highest point it is zero m per second anyways, we're not given the change in vertical displacement, we don't know it, but we're trying to find time. This is our target, oops target our target or target. So with that in mind we have these three variables. So we can use a kid a Matic equation that uses those three variables but excludes this excludes the vertical displacement and that is the equation. And let's write this in green, the equation that the final Y to the initial Y plus the acceleration in the Y direction, multiplied by T. Substituting in the values that we know. And excluding the variables. For sake of simplicity, we have Is equal to 19.9 plus negative 9.8 multiplied by T. So therefore simplifying this T is equal to 19.9 divided by 9.8 which equals 2.03. But don't be fooled. We're not done yet. This is only the time it takes to reach maximum height. This is max height. So the time it takes to reach the ground again will be twice maximum height. Because if this is roughly in the center. The time time here will also be the same as the time to the end time to the end. So we have to take tea two and multiply it by two. So two T is equal to 4. seconds. With that in mind we can go ahead and plug it back into this equation and now we can go ahead and solve so we have that delta X. The change of horizontal displacement is equal to the initial horizontal velocity multiplied by time, so this is equal to 26.4 m per 2nd, 26.4 m per second, multiplied by 4. seconds 06 seconds, giving us a final answer Of 107 m. Therefore, The distance that the seed lands at the distance from where it starts is m. I hope this helps. And I look forward to seeing you all In the next one.

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. (d) How far from its firing point does the shell land?

Watch next