(II) A fire hose held near the ground shoots water at a speed ...

Question

(II) A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3–44)? Why are there two different angles? Sketch the two trajectories.
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Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem at what angle or angles should the soccer player kick the ball so that it lands 12 m away as shown in the figure given that the ball's initial speed is 13 m per second. OK. So that's our end goal is we're trying to figure out what angle or angles should the soccer player kick the ball such that it will land 12 m away as shown in the figure provided that the ball's initial speed is 13 m per second. So that will be our final answer. Our final answer is trying to figure out what angle or angles are required in order to reach these conditions that we're asked to solve for in the problem itself. Awesome. So as we could see in our figure that's shown below, we have our soccer ball, which it's on the ground and the ground is represented by this gray bar, this gray line. So when the soccer ball is initially on the ground and then it's kicked up into the air at some angle or angles, which is gonna be denoted as theta zero or theta O and we're trying to get it to go 12 m. So basically, the ball's gonna be kicked, it's gonna go up into the air and it's gonna travel in the air until it hits the ground 12 m away. And the blue, this blue like a parabolic shape here. this blue little arch here with the arrow is representing denoting the ball's motion as it goes from the ground into the air to the ground again. Awesome. So with that in mind, we're also given some multiple choice answers and it looks like surprise. There's two angles that we're trying to solve for and they're all in units of degrees. So let's read them off to see what our final answer pair might be. A is 22 and 68 B is 28 and 62 C is 44 and 46 and D is 56 and 34. Awesome. So first off, let us note that the ball is launched from the ground and it will land on the ground again at some point. Therefore, we must recall and use the equation or the horizontal range which we denote range as capital R and it is equal to V zero squared multiplied by sine of two, multiplied by theta zero, all divided by G where G is the gravity and V zero is the launch velocity and theta zero is the launch angle. And once again, capital R is the horizontal range. So now we must rearrange the horizontal range equation to isolate and solve for theta zero, which is the launch angle. So let's start to isolate and solve for theta zero. So sine of two multiplied by theta zero is equal to capital R multiplied by G all divided by V zero squared. So now we could plug in our known variables to solve for sine of two multiplied by theta zero is equal to 12 m multiplied by gravity which is 9.8 m per second squared. Awesome. And this value is divided by our launch velocity which we are told that it is 13 m per second squared. Awesome, which this is equal to. So sine of two multiplied by theta zero is equal to 0.6959 and we round to four decimal places which we can now simplify this a little, a little bit further because we're trying to isolate and solve for theta zero. So two multiplied by theta zero is equal to the inverse sign of 0.6959 which is equal to 44 degrees when we round to the nearest hole number. So finally getting theta zero all by itself, we need to take 44 degrees and divide by two. And when we do that, we will get 22 degrees which is one of our final answers for the angle hooray, we did it. However, we must note that there will be two possible angles because both angles will provide the same range value. So the other possible angle can be found by subtracting 22 degrees from 90 degrees like so, so theta zero will be equal to 90 degrees minus 22 degrees, which is equal to 68 degrees. And that is our final answer for our second possible angle. Hooray, we did it and this corresponds with multiple choice answer letter A which is 22 degrees and 68 degrees. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

(II) A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3–44)? Why are there two different angles? Sketch the two trajectories.
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Symmetrical Launch practice set

Struggling with Physics?

(II) A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3–44)? Why are there two different angles? Sketch the two trajectories. <IMAGE>

Watch next

Symmetrical Launch practice set

(II) A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3–44)? Why are there two different angles? Sketch the two trajectories.
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