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  -> Quadratic Equations Lessons -> 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 1137679: 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 MathLover1(20849) About Me  (Show Source):
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Gloria is skeet shooting. The height of the skeet is modelled by the function
h%28t%29=-4.9t%5E2%2B32t%2B2
where h%28t%29 is the height in meters after t seconds.

The path of Gloria’s pellet is modeled by the function
g%28t%29=28.5t%2B1 , with the same units.


a) How high off the ground will the skeet be when it is hit?

when h%28t%29+=g%28t%29+, and it will be at the point where h%28t%29+and+g%28t%29+}intersects

first find the time when
h%28t%29+=g%28t%29+
-4.9t%5E2%2B32t%2B2=28.5t%2B1
0=4.9t%5E2%2B28.5t-32t%2B1-2
4.9t%5E2-3.5t-1=0
t≈-0.218733 seconds->Reject the negative solution
t≈0.933018 seconds

now find height
h%28t%29=-4.9%280.933018%29%5E2%2B32%280.933018%29%2B2
h%28t%29=-4.2655606827876%2B29.856576%2B2
h%28t%29=27.59

the point where h%28t%29 and g%28t%29+intersects is
(0.933018, 27.59)

b) After how many seconds will the skeet be hit?
after t=0.933018 seconds
it will take 0.933018+seconds to hit the skeet and the skeet will be
27.59 meters in the air