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(II) Suppose the kick in Example 7 of "Kincmatics in Two or Three Dimensions; Vectors" is attempted 36.0 $\mathrm{m}$ from the goalposts, whose crossbar is 3.00 $\mathrm{m}$ above the ground. If thefootball is dirccted perfectly between the goalposts, will it pass over the bar and be a field goal? Show why or why not. If not, from what horizontal distance must this kick be made if it is to score?

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No the football won't clear the goal post, it has to be between 4.5 and 34.7meters to clear it.

Physics 101 Mechanics

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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So here we can say that Delta X will eat be ex initial T where T would be equal to Delta X, divided by the ex initial which would simply be equal to V X because there's no no acceleration in the extraction. So this would be 36.0 meters divided by V initial. So in this case would be 20 0.0 co sign of 37 degrees. Units here would be meters per second and this would equal 2.254 seconds. At this point, we can solve for the, um, the height of the final, the maximum height of the ball. So we can say that why, rather this would be the final height of the ball. So why final equals y initial? Plus the why initial t plus 1/2 g t squared were going to say that G is gonna be negative, which means that we're choosing upwards to be positive. So we can say that why initial is zero and we can say that this is going to be equal to 20 0.0 sign of 37 degrees, uh meters per second multiplied by 2.254 seconds and then we can say plus 1/2 times negative 9.80 meters per second squared multiplied by 2.254 seconds. Quantity squared and this is giving us 2.24 meters. The field goal is three meters high, so 2.24 meters being less than three meters high. Ah, we can say that the football does not clear the bar. Attn. This point it is 0.0.76 meters too low. So in order to find the sufficient amount of a sufficient range, we can say that why final equals y initial plus B Y initial T plus 1/2 gt squared. Now here we are going to say that he can't kick the ball, uh, any harder. Um however, we we confined time. We can find the time that it must be in in the, um air so that we can find a range which with from which he can kick the football and make the field goal so we can say 3.0 meters. This is gonna equal zero. So that term is eliminated and this would be 20 sign of again 20 sign of 37 degrees multiplied by tea or rather meters per second multiplied by t and then minus 4.9 zero meters per second squared multiplied by t squared. Ah, this is a quadratic equation. As you can see. So t I's going to you consult. You can use your t 84 85 89 in order to solve for t and T is gonna equal 2.1757 seconds Uh, Delta X. But then the equal to again the x t This would be equal to 15. Or rather, we can say 20 co sign, uh, 37 degrees multiplied by 2.1 2.1757 seconds on Sorry, the second I forgot Thio mention the second answer would be 2.8 0.2814 seconds. This would be again this quadratic equations. Of course, we're gonna get to values for tea. And so the two values would indicate the range of about of the horizontal displacement from which we can kick the football. This is going to equal 34 0.746 meters and then Delta X equals again 20 co sign of 37 degrees here is gonna be multiplied by 2.2814 seconds. This is giving us 4.49 meters. So here the kick, we must be made in the range from 4.49 meters 2 34.746 meters. That is the end of the solution. Thank you for watching.

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