SOLUTION: Please help with this question.
Gloria is skeet shooting. The height of the skeet is modelled by the function h(t)=-4.9t^2+32t+2 where h(t) is the height in metres after t second
Question 1137676: Please help with this question.
Gloria is skeet shooting. The height of the skeet is modelled by the function h(t)=-4.9t^2+32t+2 where h(t) is the height in metres after t seconds. The path of Gloria’s pellet is modelled by the function g(t)=28.5t+1 , with the same units.
a) How high off the ground will the skeet be when it is hit?
b) After how many seconds will the skeet be hit? Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! set g(t) = h(t), and solve for t
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28.5t +1 = -4.9t^2 +32t +2
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-4.9t +3.5t +1 = 0
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use quadratic formula to solve for t
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t = (-3.5 +square root((3.5)^2 -4 * (-4.9) * 1 ))/(2*(-4.9)) = -0.2187
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t = (-3.5 -square root((3.5)^2 -4 * (-4.9) * 1 ))/(2*(-4.9)) = 0.933
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We ignore the negative value for t
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a) h(t) = -4.9t^2 +32t +2
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h(0.933) = -4.9 * (0.933)^2 + 32 * (0.933) + 2 = 27.5906
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The skeet will be 27.5906 metres above the ground when it is hit
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b) The skeet will be hit after 0.933 seconds by the pellet
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