Question 414.
In a game of rugby, a penalty kick is taken. The formula H(x) = 0.889x - 0.0198x^2 models the flight of the rugby ball. H(x) measures the height of the ball above the ground (in metres) at a distance x metres along a straight line on the ground beginning at the point where the penalty kick is taken and extending beyond the two goal posts.
i) How far from where the ball was kicked did it land (nearest meter)?
ii) Calculate the maximum height H reached by the ball (nearest meter).
iii) After the kick is taken, it just clears the cross-bar by 1cm. How far away from the goal posts does the formula suggest the penalty kick was taken? Set up the equation to be solved, and the equation that will help you solve it. Round sensibly.
Note: The height of the cross-bar is 3 m above the ground.
Answer 414.
i)
H = 0 where the ball lands
=> 0 = 0.889x - 0.0198x^2
=> 0 = 0.889 - 0.0198x ... [because x is not zero where it lands]
=> x = 45 m.
ii)
Maximum height is reached at x = (1/2) * (0.889/0.0198)
=> H = 0.889 * (1/2) * (0.889/0.0198) - (0.0198) * (1/2)^2 * (0.889/0.0198)^2
=> H = (1/2) (0.889)^2/(0.0198) m
=> H = 20 m.
iii)
Plugging H = 3.01
=> 3.01 = 0.889x - 0.0198x^2
=> x^2 - (0.889/0.0198)x + (3.01/0.0198) = 0
=> x^2 - 44.899x + 152.02 = 0
=> x = (1/2) [44.899 ±√[2015.92 - 608.08]
=> x = (1/2) (44.899 ± 37.521)
=> x = 3.689 m or 41.21 m.
41.21 m is the likely answer.
Link to YA!
In a game of rugby, a penalty kick is taken. The formula H(x) = 0.889x - 0.0198x^2 models the flight of the rugby ball. H(x) measures the height of the ball above the ground (in metres) at a distance x metres along a straight line on the ground beginning at the point where the penalty kick is taken and extending beyond the two goal posts.
i) How far from where the ball was kicked did it land (nearest meter)?
ii) Calculate the maximum height H reached by the ball (nearest meter).
iii) After the kick is taken, it just clears the cross-bar by 1cm. How far away from the goal posts does the formula suggest the penalty kick was taken? Set up the equation to be solved, and the equation that will help you solve it. Round sensibly.
Note: The height of the cross-bar is 3 m above the ground.
Answer 414.
i)
H = 0 where the ball lands
=> 0 = 0.889x - 0.0198x^2
=> 0 = 0.889 - 0.0198x ... [because x is not zero where it lands]
=> x = 45 m.
ii)
Maximum height is reached at x = (1/2) * (0.889/0.0198)
=> H = 0.889 * (1/2) * (0.889/0.0198) - (0.0198) * (1/2)^2 * (0.889/0.0198)^2
=> H = (1/2) (0.889)^2/(0.0198) m
=> H = 20 m.
iii)
Plugging H = 3.01
=> 3.01 = 0.889x - 0.0198x^2
=> x^2 - (0.889/0.0198)x + (3.01/0.0198) = 0
=> x^2 - 44.899x + 152.02 = 0
=> x = (1/2) [44.899 ±√[2015.92 - 608.08]
=> x = (1/2) (44.899 ± 37.521)
=> x = 3.689 m or 41.21 m.
41.21 m is the likely answer.
Link to YA!
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