Question

Answered step-by-step

Problem 79 Hard Difficulty

A placekicker is about to kick a field goal. The ball is 26.9 ${m}$ from the goalpost. The ball is kicked with an initial velocity of 19.8 ${m} / {s}$ at an angle $\theta$ above the ground. Between what two angles, $\theta_{1}$ and $\theta_{2},$ will the ball clear the $2.74-{m}$ -high crossbar? (Hint: The following trigonometric

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Griffin Goodwin

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Physics 101 Mechanics

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Video Transcript

So the question states that a fuel goal kicker can kick a ball at 19 0.8 meters per second and there is a field goal that is 26.29 26.9 meters away with the height of 2.74 meters. Uh, tall and we're trying to find at what two angles can the ball be kicked at so that it passes just threw the upright of the the field goal. So to do this, the first thing we should do is figure out a way to represent the time in terms of the theater angle. So we know that the horizontal component of this, um velocity vector is going to be so Visa Becks is going to be 19.8 times co signed data and we know this because coastline data is adjacent overhype on news where the adjacent is visa Bendixen the hypothesis 19.8 So we can rearrange it and get this equation. Now that we know V of X, we can use our Kinnah Matic equations which state that the velocity in the X direction multiplied by the time should be eagle toothy range of the projectile so we know that the velocity is 19.8 times co sign of data Times T should be equal to 26.9 meters, which means t the time that the project I was in the Air Force should be 26.9 divided by 19.8 times co sign of theater Now that we know this, we can use another Kinnah Matic equation too eso for the state of value that we want. So the one we're gonna use now is the one that states that the change in displacement in the Y direction is equal to the initial velocity in the Y direction times the time plus 1/2 times the acceleration times the time squared. So we know the change in displacement is going to be 2.74 meters. We know the initial velocity in the Y direction is 19.8 times signed the data. It's just the same thing. Similar calculation to the horizontal of lost e calculation. So in 19.8 time sign data Times T, which we calculated up here to be 26.9, divided by 19.8 times co sign of data plus 1/2 times the acceleration due to gravity, which is negative 9.8 meters per second squared times This time square, it's a 26.9 over in 19.8 co sign data and all this will be squared. So now that we have this, we can simplify it a little bit, so we're gonna have 2.74 is equal to. So this signing co sign will become tangent of Fada and these 19.8 will cancel out. So we'll just be left with 26 0.9 times the tangent of state. Uh, plus, um, 1/2 times negative, 9.8 times 26.9 squared over 19.8 squared times, one over co sine squared, which the same thing as seeking squared, which is also the same thing as one plus 10 tension square. So we'll have one plus tangent squared here of data. So now that we have this, we can simplify it a little bit further and get a, um, equation that's equal to zero. So when we I'm starting from the right and side here, so we're gonna have one plus 10 squared the, uh, times this number, which I simplified the one this 1/2 blah, blah, blah, blah, blah. It's going to be simplified into negative 9.0 for for to plus 26.9 times the attention of data minus 2.74 It's gonna be eagle Zero and we can distribute this one plus 10 square data. And when we do this, we will get the following equation. Get nine negative, 9.0 for for two can squared. They'd, uh, plus 26.9 10 data minus 11.7842 is equal to zero. And at this point, we can set our 10 of Fada equal to X, and we can solve it as a, uh, using quadratic formula to find the roots. Or we can just use a graphing calculator to find the roots. And so when we use a graphing calculator, we'll see that one of the routes is 10.534 and the other is 2.44 So tan of data will either eagle 0.534 or 2.44 So now if we want to find the angle fatal, we can just take the 10 inverse of both of these, so 10 in verse of 0.534 as well as the 10 inverse of 2.4 for And we'll get that thief state of values that work. Um, for this for the situation is 28 0.1 degrees as well as 67.71 degrees, and that's your final answer.

Get More Help with this Textbook

John D. Cutnell, Kenneth W. Johnson

Physics

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Physics

Related Topics

Discussion

Top Physics 101 Mechanics Educators

Liev Birman

Marshall Styczinski

Zachary Mitchell

Jared Enns

Physics 101 Mechanics Courses

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

John D. Cutnell, Kenneth W. Johnson

Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Jared Enns

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Jared Enns

John D. Cutnell, Kenneth W. JohnsonPhysics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Jared Enns

John D. Cutnell, Kenneth W. Johnson

Physics