SOLUTION: 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 i

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 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 i      Log On


   

Question 1137659: 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 josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Find the intersection of the 2 equations
+h%28t%29+=+g%28t%29+
+-4.9t%5E2+%2B+32t+%2B+2+=+28.5t+%2B+1+
+-4.9t%5E2+%2B+3.5t+%2B+1+=+0+
+t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+t+=+%28-3.5+%2B-+sqrt%28+3.5%5E2-4%2A%28-4.9%29%2A1+%29%29%2F%282%2A%28-4.9%29%29+
+t+=+%28-3.5+%2B-+sqrt%28+31.85+%29%29%2F%28-9.8%29%29+
+t+=+%28+-3.5+-+5.64358+%29%2F%28-9.8%29+%29+
+t+=+9.1436%2F9.8+
+t+=+.933+ sec
and
+g%28.933%29+=+28.5%2A.933+%2B+1+
+g%28+.933+%29+=+26.591+ m
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The pellet is 26.591 m off the ground when it is hit
It is hit .933 sec after firing
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Here's the plot:
+graph%28+600%2C+400%2C+-1%2C+4%2C+-6%2C+60%2C+-4.9x%5E2+%2B+32x+%2B+2%2C+28.5x+%2B+1++%29+
Looks about right