You're 6.0 m from one wall of the house seen in FIGURE P4.55. You...

Question

You're 6.0 m from one wall of the house seen in FIGURE P4.55. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground.c. At what angle should you toss the ball?

Accepted Answer

Hey, everyone. Welcome back in this problem. We have a thief that's gonna steal some diamonds in a museum. Security is gonna learn about this and carry out a search operation. The thief ends up locked inside a wall as the entrances and exits are sealed. And in order to save himself from getting caught, he's going to throw the diamond pouch to his partner outside of the gate. And we're asked to determine the angle at which the poach needs to be thrown to safely reach the partner outside of that gate. And we're given a diagram of what's going on here. So we have our thief inside with the diamond pouch. The thief throws it from a height of 2 m all the way to the partner thief outside. There's a wall in between them, 4 m high, 1 m wide. And the poach is going to travel over that in additional 0.5 m and the wall is 5 m from each thief respectively. We have four answer choices. Option A 42.3 degrees. Option B 21.8 degrees. Option C 68.2 degrees and option D 39.3 degrees. So let's get started and we're gonna get started with the throw that goes to the other partner, the, OK. When we wanna figure out the angle, what we need to do is figure out the velocity and we need to know the initial velocity in the X and Y direction that this coach is getting thrown at in order to determine the direction. OK. So let's look at our variables in the throw to the part. OK? And we're gonna do that in bullet and when this is thrown to the partner in the X direction and we don't know the velocity or the speed VX, we know delta X. OK. From our diagram, the thief is 5 m from the wall, the wall is 1 m thick and then the second partner is 5 m from that outer edge. And so in total, the poach has to go 11 m in the horizontal direction and we don't know how long this takes. OK. So in terms of figuring out this X velocity, we can't do it from this information alone. We only have one piece of information here. If we switch over to the Y direction, we have our initial velocity in the Y direction. V. Not Y, we don't know what that is either, but we wanna find out we don't know the final velocity. When the partner catches it. The change in position. Delta Y in the vertical direction is just going to be 0 m. Hey, we're gonna assume that the two partners are about the same height and so the partner is catching it at the same height that it was thrown from. OK. So that change in vertical position is zero. We have acceleration due to gravity acting, we're gonna take upwards into the right to be our positive directions. And then our acceleration is going to be negative 9.8 m per second squared. And again, we don't know how long this is. So looking at just this information, we also can't solve for anything. We only have two known values. Remember we need at least three to solve a kinematic equation. So what else can we do? Well, we're given information about the peak. OK, this maximum height that the poach reaches in our diagram. And we also know some information about speed at the maximum height. And remember that the vertical speed will be zero as it changes direction momentarily and then or that velocity. So let's switch over and we're gonna do this in green and we're gonna look at throwing it to the maximum height instead of all the way to the other partner. So to the maximum height, what does the throw look like? OK. Now the initial velocity in the X, we don't know and we still don't know how fast this is being thrown. The horizontal displacement. Well, the maximum occurs right at the center of the wall. So we have 5 m to the wall and then half of that thickness of the wall, the wall is 1 m, half of that is 0.5 m. And so the horizontal distance is going to be 5.5 m. And again, we don't have any information about time switching over to the vertical direction again. All right, we're just getting everything laid out. So we know what we need to solve that initial velocity. We still don't know what's interesting here though is this throw, whether we stop and look at what's happening at the middle or we stop and look at what's happening when it gets all the way to the partner. Those initial velocities are the same in both cases. OK? So the VX and the VNY in blue are the same as the VX and VNY in green, the final velocity in the vertical direction. We've talked about this. We're at the maximum height, the velocity is going to be zero, that poach is going to come to rest momentarily as it changes direction. And so that velocity is gonna be 0 m per second. The change in vertical position. Well, the pouch is launched from 2 m in the air. The height of the wall is 4 m. The pouch goes an additional 0.5 m, ok? So the pouch is going to be at 4.5 m measured from the ground, but it starts at 2 m from the ground. So we have 4.5 m minus 2 m, which tells us that the vertical displacement is 2.5 m. OK. Again, we have our acceleration due to gravity negative 9.8 m per second squared. And we don't know anything about the time. T now, what we can see is that we have three known values in this green situation. OK. When we're talking about getting to the maximum height, that means that we can solve our kinematic equation. And what variable should we try to find? Right. Well, what we're gonna try to find is V not Y remember that initial velocity, whether we're talking about looking just till the maximum height or till the partner thief is the same. OK? That launch speed, launch velocity is the same. So what we're gonna do is use this vertical direction. Look for VNY, the equation we're gonna use, we're gonna use the three variables. We know VF delta Y and A and the one that we're looking for V not and we get the following VFY squared is equal to V, not Y squared plus two multiplied by a multiplied by delta Y substituting in our values, we have zero is equal to V not Y squared, close to multiplied by negative 9.8 meters per second squared, multiplied by 2.5 m. We're gonna move all of our numbers to the left hand side and then just rewrite this. So we have V not Y squared. Is going to be equal to 49 m squared per second squared, which gives us an initial velocity in the Y direction of 7 m per second. Good. All right. Now, we're gonna put a red box around that because we don't wanna lose track of it. So what we figured out now is the Y velocity that this pouch is thrown. OK. That means that we have that Y velocity for other situation as well. Remember what we're trying to find. We're trying to find the launch angle. And in order to find the angle that, that poach needs to be thrown, we needed the velocity, the Y direction and the X direction. Right now, we only have the Y direction. OK. Going back to our blue equations, we need to find VX. In order to find VX, we have to know the time T but guess what? We can calculate that. Now, we now have three known values in the vertical direction. We can use those, calculate the time T, then we will have T in order to calculate the X. So let's go ahead and do that. OK. So this is a puzzle piece. That's why we write out all the information at the beginning. It helps you see what you have, what you need. And then you can work through step by step how to solve. So we're gonna go ahead, we're gonna use the equation that has the variables we know. And the one that we're looking for, we have delta Y is equal to V not Y multiplied by T plus one half, multiplied by A Y multiplied by T squared. Substituting in our values 0 m is equal to 7 m per second, multiplied by T plus one half multiplied by negative 9.8 m per second squared multiplied by T squared. On the right hand side, we're gonna factor out A T. So we factor T and we're left with 7 m per second minus 4.9 m per second squared, multiplied by T, we have two things multiplied together that are equal to zero. That's gonna be true if either of those factors are equal to zero. So from the first factor we get that T is equal to zero. That's just our initial point. OK? Initially, as the ball is, or the pouch is just about to be thrown, that's that time point there. They were at the same height as when the partner thief catches it. So we wanna find that other time point when the partner thief catches it after time zero. So what we do is set the second factor to zero, 7 m per second minus 4.9 m per second. Squared T is equal to zero, tells us that 4.9 m per second squared, multiplied by T is equal to 7 m per second and dividing by 4.9 m per second squared on both sides is this that the time is about 1.42857 seconds. Again, putting that in a red box. So we don't lose track of it. And now we found the time it takes for the pouch to go from the thief on the inside all the way to the thief on the outside. Let's go back up. Take a look at our information. Remember we're trying to find VX. We now have the time, T 1.428 57 seconds. OK? So we have delta X, we have T we can solve for VX. Let's go ahead and do that now. OK. So we're called VX is gonna be delta X divided by T and the change in position divided by the change in time. So we have VX is 11 m divided by 1.42857 seconds, which gives us an X velocity of about 7.7 m per second. All right, we're getting really, really close. We now have our initial velocity in both directions. We know the X direction and Y direction of that initial velocity and we can use that to find the angle. So let's go ahead and draw a little diagram of what we have going on. So we have an X velocity pointing to the right and the positive direction with a magnitude of 7.7 m per second. And then we have a worry velocity pointing upwards the positive direction with a magnitude of 7 m per second. And the actual throw is going to be along the hypotenuse of the triangle that that makes and this is our angle theta between the two. That is what we are looking for. Now, we have a right angle triangle. We can use our trigonometric ratios. We want to relate the opposite in adjacent sides. So we're gonna use tangent, we have the tangent of theta is going to be the opposite side, which is going to be the Y provided by the adjacent side which is going to be the X. Hey, this tells us that tan theta is going to be 7 m per second divided by 7.7 m per second. You can take the inverse tangent to find theta. We have theta is the inverse tangent of seven divided by 7.7. Those units will divide out and we get an angle of approximately 42.27 degrees. Hm All right. So there is a lot of steps there. The main thing, write out all the information we have and then just piece by piece, solve for what we can and getting towards the answer. So we needed to find the horizontal and vertical velocities in order to find the angle that, that poach was thrown at comparing this to our answer choices, rounding to one decimal place. We can see that the correct answer here is option a 42.3 degrees. Thanks everyone for watching. I hope this video helped see you in the next one.

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You're 6.0 m from one wall of the house seen in FIGURE P4.55. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground.c. At what angle should you toss the ball?

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