# Video: Analyzing the Dynamics of an Object in Both Linear and Ballistic Motion

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m above the muzzle of the mortar. Neglecting air resistance, calculate the shell’s velocity when it leaves the mortar. The mortar itself is a tube, 0.450 m long. Calculate the magnitude of the average acceleration of the shell in the tube as it goes from zero to the velocity at which it leaves the mortar.

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### Video Transcript

A 2.50-kilogram fireworks shell is fired straight up from a mortar and reaches a height of 110 meters above the muzzle of the mortar. Neglecting air resistance, calculate the shell’s velocity when it leaves the mortar. The mortar itself is a tube, 0.540 [0.450] meters long. Calculate the magnitude of the average acceleration of the shell in the tube as it goes from zero to the velocity at which it leaves the mortar.

We have a two-part problem here where we’re asked to solve first for the shell’s velocity when it leaves the mortar, and secondly what the average acceleration of the shell was while it was in the mortar. We’ve been given some helpful information we wanna keep track of. First, we know the mass of the shell, 2.50 kilograms. And we know that it reaches a maximum height of 110 meters above the muzzle of the mortar. And finally, we know how long the mortar is; its 0.450 meters.

Let’s address this first question of the shell’s velocity when it leaves the mortar using some of this information we’ve been given. As a first step here, we can draw a diagram of the situation. So we have our mortar pointed straight up and our fireworks shell is released from the mortar, goes up 110 meters, and then vertically comes to rest, so that the velocity of the shell at that point is zero. Now we’re told the mass of the shell. And what we wanna figure out is, what is this initial velocity, the velocity of the shell as it leaves the muzzle so that it attains this maximum height of 110 meters.

This is a projectile motion problem. And as such, we can helpfully rely on the kinematic equations to let us find out what this initial velocity is. So let’s review those equations now. These kinematic equations apply when a projectile experiences constant acceleration like in this case with gravity. Now let’s look through this list and see if we find an equation that matches what we’ve been given in the problem and also includes the variable that we want to solve for. Scanning down from the top, we see that this second equation, that the final velocity of an object squared equals the square of its initial velocity plus two times its acceleration times the distance it travels, is a match for our scenario.

Let’s map the variables in this equation to our own particular situation. 𝑣𝑓 squared or 𝑣𝑓 is the final velocity of our projectile. Since we’ve decided that that would be where the fireworks shell is at its peak, that final velocity in our case is zero. So this first term drops out. 𝑣𝑖 is the variable we wanna solve for. 𝑎 is the acceleration, in this case, due to gravity. We’ll represent that as a lower case 𝑔. And 𝑑 is our maximum vertical distance the shell travels. We’re given that to be 110 meters. So this equation is a great match for what we wanna solve for in the variables we’ve been given. Let’s now rearrange it to solve for 𝑣 sub 𝑖, the initial velocity of the shell as it leaves the muzzle.

We can take both sides of this equation and subtract two 𝑎𝑑 from each side. In that case, the two times 𝑎 times 𝑑 on the right-hand side of our equation cancels out. Now we take the square root of both sides which will lead to the cancelling of the squared term above 𝑣 sub 𝑖 and the square root itself. So we’re left with an equation that reads: 𝑣 sub 𝑖 equals the square root of negative two 𝑎𝑑. Now we can plug in for 𝑎, the acceleration, and 𝑑, the distance. First, let’s establish a convention of which direction in our case is positive. Let’s choose up to be positive. That means that motion in that direction will have a positive sign to it and motion downward will have a negative sign.

Now the acceleration that this shell experiences is acceleration due to gravity that points down. So when we plug in for 𝑎, we’ll use negative 9.8 meters per second squared. Now that that’s in, let’s write in our third and final term 𝑑, the distance that the shell travels above the muzzle, 110 meters. When we multiply these three numbers together and take their square root, the result we find is 46.4 meters per second. That’s the velocity that the shell begins with when it leaves the muzzle of the mortar. So that’s part one of our problem. We found the initial velocity of the shell just as soon as it’s fired out of the mortar.

Now let’s move on to solving part two, where we want to know what is the average acceleration the shell experiences while it’s inside the mortar. So we have our answer to the first part of this problem, the initial velocity, and we’ll keep that up top. And now we wanna answer the question what is the average acceleration, we’ll call it 𝑎 sub avg, that the shell experiences while it’s being accelerated in the muzzle. And here is where we can use a little bit more of the information we were given in our original problem statement. We were told in that original problem that the muzzle has a length of 0.450 meters. So we can draw that in on our diagram. There we go, we’ve included that in our diagram. And let’s call that distance 𝑙, for shorthand.

Now to make a headway with this problem, let’s recognize that we’re looking for an average acceleration, which means that we can effectively assume a constant acceleration throughout the whole time that the shell’s accelerating through the mortar. Now when you hear constant acceleration, then we enter into the world of kinematic equations. But before we revisit our list of equations, let’s just describe a little bit what’s going on here between the time that the shell is at rest in the muzzle and the time that it leaves the muzzle. So here we have a-a close up view of our mortar. And we have two times, we’re calling them the initial time, 𝑡 sub 𝑖, and the final time, 𝑡 sub 𝑓. At 𝑡 sub 𝑖, that initial time, our fireworks shell is at rest in the bottom of the mortar, and it’s ready to be accelerated by the ignition and explosion of the gases coming out of the mortar. So at that point, the shell has an initial velocity of zero meters per second. Now don’t confuse this 𝑣 sub 𝑖 with the previous 𝑣 sub 𝑖, which is actually the speed of the shell when it leaves the muzzle. So we’ve rescaled the time we’re talking about. And now our initial velocity is zero because we’re referring to the time when the shell was at the bottom of the muzzle.

Okay. So the shell is at the bottom of the muzzle at rest. Then the muzzle gunpowder is ignited. It explodes and forces the shell up and out of the muzzle. When the shell leaves the muzzle at the final time, it has acquired a velocity we’re calling the final velocity, 𝑣 sub 𝑓. And we solved for that final velocity in part one of this problem. So the 𝑣 sub 𝑓 in this part, part two of the problem, is equal to 𝑣 sub 𝑖 in part one. In other words, 𝑣 sub 𝑓 equals 46.4 meters per second. Great! So now we have our setup more clearly laid out. We know the length of our mortar, we know the initial speed that the shell had in the mortar, we know its final speed, and we wanna know what is the average acceleration this shell experienced as it moved through this mortar.

So now we’re ready to go back to our kinematic equations and see if we find a good match for this part two of our scenario. And in fact we find that the same equation we used earlier is a perfect fit for what we wanna find. Our final velocity is a known value, that’s 46.4 meters per second. Our initial velocity is known, that’s zero because our shell is at rest at first. The acceleration 𝑎 is what we’re looking to solve for. We’ve labelled that 𝑎 sub avg, that’s what we’re looking to solve for. And 𝑑, the distance, is in our case 𝑙, the length of the muzzle, which is given as 0.450 meters. So we’re now ready to use this simplified version of the equation to solve for the average acceleration that the shell experiences.

So here is that equation with the variables switched out to represent the variables that we’re working with, 𝑎 sub avg and 𝑙. Now we wanna solve for 𝑎 sub avg. So to do that, let’s divide both sides of the equation by two times 𝑙. When we do that, the two and the 𝑙 on the right-hand side of our equation cancel out, and we’re left with a final expression that reads: 𝑎 sub avg equals 𝑣𝑓 squared over two 𝑙.

Now we’re ready to plug in the values we know. 𝑣 sub 𝑓 is 46.4 meters per second and 𝑙 is 0.450 meters. When we enter these numbers into our calculator, we end up with an average acceleration of 2.40 times 10 to the third meters per second squared. That’s the average acceleration that our shell experiences over this speeding-up process from being at rest in the bottom of the muzzle to leaving the muzzle with its maximum velocity.

And that completes the second part of this two-part question.