Physics - Mechanics: Projectile Motion (4 of 4)

by: Michel van Biezen

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[0.03]
here's another example of how to do projectile motion problem let's read the problem see what it's all about a basketball player attempts to make a basket from a distance of 8 meters the basket is 3 meters high and the ball leaves her hands from a height of 1 meter if she throws the ball at an angle of 45 degrees what does the initial velocity of the ball need to be all right usually a problem like this strikes the fear into a lot of students it used to for me as well Wow how do you start a problem like this where do you begin well it's not a bad idea to begin by writing down all the things that are given so let's go find them it says here that the distance to the basket is 8 meters so we can say x equals 8 point zero zero zero meters the basket is three meters high so Y or H is equal to three point zero zero Mears we're given this the ball leaves our hands from a height of 1 meter so that would be the initial height the starting height would be one point zero zero meters and she throws the ball at an angle of 45 degrees so we have this one now we go over here say the angle theta is equal to 45 degrees and finally they say what should be the initial velocity to attain that basket so V initial equals question mark that's what's being asked all right so we have a pretty good idea now what's given and what they're asking for but again in order to really feel what we need to do here try to figure out what we need to do we want to make a nice little graph or drawing of what's happening so here's our basketball player throwing the basketball at an angle of 45 degrees theta equals 45 degrees with some initial velocity which we don't know that's what they're asking for all right the basketball will move in a parabolic path and we're trying to hit the basket the basket has a height of three meters that's the final height why right there and they also tell us that the basket is eight meters away x equals eight meters away like so okay that gives us a much better feel for what's happening here's a basketball player shooting the basket or shooting the ball to the basket right they're supposed to hit the basket eight years away baskets three meters high the ball leaves the hands of the player at a height of 1 meter so y-initial equals 1 meter this would be H or y equals 3.0 meters while it would be the final height all right what's next what's the strategy we need a good strategy and I would say projectile motion time in the air probably not a bad way to start time in the air now since we're given the distance I'm going to use the horizontal component of the velocity to find time in the air oh but I don't have the horizontal component yet so let's do that next we need the horizontal component of the initial velocity and the vertical component of the initial assi so V initial in the X Direction is equal to V initial times D since the components of jacent to the angle times the cosine of theta and over here for the vertical component V initial in the Y Direction is equal to V initial times the sine of theta plug it in what those are so it's equal to oh we don't know what that is that's V initial times the cosine of 45 degrees and the cosine of 45 degrees is 0.707 so we can say that this is equal to zero point 707 V initial and since it's exactly the same for the sine of theta because at the angle of 45 degrees the sign in the cosine are exactly the same we could say that this is equal to zero point 707 times V initial all right so now that we've expressed the philosophy Nishal for the X and y-component we can calculate a time in the air so we could say that x equals x sub naught plus V sub naught in the x-direction times time plus 1/2 86 that we have right up here notice that in the horizontal direction the moment the ball leaves the player's hand there's no forces involved so there's no acceleration in the X direction so that's zero we can assume that the initial position x equals zero at time equals zero so this equation becomes x equals V initial in the X Direction times time if I not plug in everything that we know or my let's solve for time first let's say time is equal to X divided by V initial in the X direction I simply divide both sides the equation by V initial in the X direction and so that means that distance of 8 meters divided by Venus on the x-direction which is zero point 707 times V initial so here I have now expressed the time in the air in terms of the initial velocity if I now plug this into my second equation kinematics for the vertical component I'm now going to grab my second equation right here which is y equals y sub naught plus V sub naught in the Y Direction times time plus one-half GT squared remember in the vertical direction since there's gravity we have an acceleration due to gravity G which is a minus 9.8 meters per second squared I am now going to plug this value for T both in here and in here when I do that I get the following final height 3 meters initial height 1 meter I'll put parentheses around it's a little cleaner plus the initial velocity in the Y direction the initial velocity in the Y direction we have over here which is zero point 707 times the initial velocity multiplied times the time which I have over here in terms of V initial so it would be eight point Oh meters divided by 0.7 Oh 7 v initial all right now I still have a plus 1 half times G which is minus 9.8 meters per second squared times the x squared and the time is over here so I plug that in 8.00 meters divided by 0.7 oh 7 V initial and the whole thing is squared all right now notice I now end up with one single equation that only has a single unknown in the V initial and nicely here this 0.707 V initial cancels out with this 0.707 V initial so now I have 3 meters equals 1 meter and then plus 8 meter so I can move everything over to one side and have a 3.0 meters minus 1.00 meters when I bring this one across and what I have left over here is a minus 8.0 meters when I bring it to the other side equation equals 1/2 times 9.8 is a minus 4.9 meters per second squared and multiply times 8 squared which is 64.0 meters squared divided by this number squared is 0.5 times V initial squared okay so I squared this term right here which is 64 0.707 squared is 0.5 V initial squared is V squared and down one half times 9.8 is minus 4.9 combining what's on the left side I have a three minus one is two minus eight is a minus 6.00 meters equals minus 4.9 meters per second squared times 64 divided by 0.5 to 128 so that's 128 meters squared divided by V initial squared okay now I'm ready I have a negative on both sides second multi both sides by a negative one that make that positive I cannot move the V initial squared over here and the six meters down here so let me move over here so I have a little bit more room so I'm moving my V initial squared to the other side diagonally across I have V initial squared is equal to I still have a four point nine meters per second squared I still have a hundred and twenty eight meters squared and have a bring the six meters across over here divide both sides by six meters I get 6.00 meters okay so this meter cancels out this meter and I have a meter squared in the second squared and have V initial squared on that side so I can take the square root of both sides so V initial is equal to the square root of four point nine times one over seconds squared times 128 that's meters squared all divided by 6.00 and now let me grab my calculator okay 128 times four point nine divided by six and take the square root of that I get ten point two two meters or the initial equals ten point two meters that should be the initial velocity of the basketball if she has any aspirations to hit that basket and who yes I forgot something Thank You m/s yeah velocity in terms of meters wouldn't be very good with it ten point two meters



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for more math and science lectures! In this 4 lecture series I will show you how to solve different physics problems that deal with projectile motion. Problem Text: A basketball player attempts to make a basket from a distance of 8.00m. The basket is 3.00m high and the ball leaves her hands at a height of 1.00m. If she throws the ball at an angle of 45 degrees, what does the initial velocity of the ball need to be?
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