Question

Justin is skeet shooting. The height of the skeet is modeled by the equation h= -5t^2 +32t + 2, where h represents the height in meters and t seconds after the skeet is released. The path of Justin's bullet is modeled by the equation h= 31.5t + 1, with the same units. How long will it take for the bullet to hit the skeet? How high off the ground will the skeet be?

Answer 1

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 = $\frac{-.5 +/- \sqrt{.25 - 4(-5)(1)} }{2(-5)}$ (substitute)

t = $\frac{-.5 +/- \sqrt{.25 + 20} }{-10}$ (multiply)

t = $\frac{-.5 +/- \sqrt{20.25} }{-10}$ (add)

t = $\frac{-5 +/- 4.5}{-10}$ (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.