SOLUTION: 20. The height of a football is given β„Ž(𝑑)=βˆ’4.9𝑑2+18𝑑+0.8 where β„Ž(𝑑) is the height (in metres) and 𝑑 is the time (in seconds) from when the punter kicked

Algebra ->  Test -> SOLUTION: 20. The height of a football is given β„Ž(𝑑)=βˆ’4.9𝑑2+18𝑑+0.8 where β„Ž(𝑑) is the height (in metres) and 𝑑 is the time (in seconds) from when the punter kicked      Log On


   

Question 1167023: 20. The height of a football is given β„Ž(𝑑)=βˆ’4.9𝑑2+18𝑑+0.8 where β„Ž(𝑑) is
the height (in metres) and 𝑑 is the time (in seconds) from when the
punter kicked the ball. The height of a blocker’s hands is modelled by
the equation 𝑔(𝑑)=βˆ’1.43𝑑+4.26. Can the blocker knock down the
punt? If so, at what point will it happen?
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.

Regarding this problem, notice that both these formulas describe the projection of the position to the vertical direction.


They do not describe a trajectory of the ball, which is usually a parabola in (x,y)-plane.


In football, it is a rare case, when the ball moves vertically up --- so the question and the problem itself are not very realistic.


If, nevertheless, these circumstances do not embarrass you,  then consider this equation


    -4.9*𝑑^2 + 18𝑑 + 0.8 = -1.43t + 4.26


and solve it for t.


Solving this equation is a mechanical work, so I leave it to you.

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In my personal view, the posed problem is not adequate to reality.