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When a field goal kicker kicks a football as hard as he can at 45° to the horizontal, the ball just clears the 3-m-high crossbar of the goalposts 45.7 m away. (a) What is the maximum speed the kicker can impart to the football? (b) In addition to clearing the crossbar, the football must be high enough in the air early during its flight to clear the reach of the onrushing defensive lineman. If the lineman is 4.6 m away and has a vertical reach of 2.5 m, can he block the 45.7-m field goal attempt? (c) What if the lineman is 1.0 m away?

Respuesta :

Answer:

Part a)

[tex]v = 21.9 m/s[/tex]

Part b)

[tex]y = 4.17 m[/tex]

So he will not able to block the goal

Part c)

[tex]y = 0.98 m[/tex]

yes he can stop the goal

Explanation:

As we know by the equation of trajectory of the ball

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 3 m[/tex]

[tex]x = 45.7 m[/tex]

[tex]\theta = 45 degree[/tex]

now from above equation we have

[tex]y = 45.7 tan 45 - \frac{9.81(45.7)^2}{2v^2cos^245}[/tex]

[tex]3 = 45.7 - \frac{20488}{v^2}[/tex]

[tex]v^2 = 479.81[/tex]

[tex]v = 21.9 m/s[/tex]

Part b)

If lineman is 4.6 m from the football

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 4.6tan45 - \frac{9.81(4.6^2)}{2(21.9^2)cos^245}[/tex]

[tex]y = 4.17 m[/tex]

So he will not able to block the goal

Part c)

If lineman is 1 m from the football

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 1tan45 - \frac{9.81(1^2)}{2(21.9^2)cos^245}[/tex]

[tex]y = 0.98 m[/tex]

yes he can stop the goal