Answer:
0.05 seconds; 2.575 m off the ground
Step-by-step explanation:
This is a system of equations that you need to solve for the variables t and h.
h = -5t² + 32t + 2
h = 31.5t + 1
Since h will be the same when the bullet and skeet collide, then h is the same in both equations and can be set equal to each other and then you can solve for t.
-5t² + 32t + 2 = 31.5t + 1
-5t² + 32t + 2 - 31.5t - 1= 31.5t + 1 - 31.5t -1 (move everything to the left side)
-5t² + 0.5t + 1 = 0 (combine like terms)
Now we can use the quadratic formula to solve for the time.
t = [tex]\frac{-.5 +/- \sqrt{.25 - 4(-5)(1)} }{2(-5)}[/tex] (substitute)
t = [tex]\frac{-.5 +/- \sqrt{.25 + 20} }{-10}[/tex] (multiply)
t = [tex]\frac{-.5 +/- \sqrt{20.25} }{-10}[/tex] (add)
t = [tex]\frac{-5 +/- 4.5}{-10}[/tex] (square root)
t = -0.5/-10 or -9.5/-10 (combine like terms in each situation)
t = 0.05 or .95 (divide)
Since it will only hit it once, the first time it hits is the amount of time, therefore it takes .05 seconds to hit the skeet.
Next, substitute t into one of the equations to find the height at which it occurs.
h = 31.5(0.05) + 1 (substitute)
h = 1.575 + 1 (multiply)
h = 2.575 (add)
Therefore the skeet will be 2.575 meters off the ground.
