The height of a baseball (h) in meters (t) seconds after it has been tossed out a window is given by the equation h= -5t^2 + 20t + 15. A boy shoots at the baseball with a trajectory of h= 3t +3. Will the paintball hit the baseball? If so at what height and time?

well, when do you have

-5t^2 + 20t + 15 = 3t + 3?

once you have t, then plut it in to get h.

erewrg

To determine if the paintball will hit the baseball, we need to find the point at which the height of the paintball trajectory is equal to the height of the baseball.

Set the two equations equal to each other and solve for t:

-5t^2 + 20t + 15 = 3t + 3

Rearrange the equation:

-5t^2 + 17t + 12 = 0

To solve this quadratic equation, you can either factor it or use the quadratic formula. In this case, let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Using the values from the equation -5t^2 + 17t + 12 = 0, we have:

a = -5, b = 17, c = 12

Substituting these values into the quadratic formula, we have:

t = (-17 ± √(17^2 - 4(-5)(12))) / (2(-5))

Simplifying further:

t = (-17 ± √(289 + 240)) / (-10)

t = (-17 ± √529) / (-10)

t = (-17 ± 23) / (-10)

This gives us two possible values for t:

t = (-17 + 23) / (-10) = 0.6 seconds
t = (-17 - 23) / (-10) = 4 seconds

To determine the height at these times, substitute the values of t back into either of the original equations. Let's use the equation for the baseball's height:

h = -5t^2 + 20t + 15

For t = 0.6 seconds:

h = -5(0.6)^2 + 20(0.6) + 15
h = -1.8 + 12 + 15
h = 25.2 meters

For t = 4 seconds:

h = -5(4)^2 + 20(4) + 15
h = -80 + 80 + 15
h = 15 meters

Therefore, the paintball will hit the baseball at two different times at heights of 25.2 meters and 15 meters.

To determine whether the paintball will hit the baseball, we need to find the time when both trajectories intersect. In other words, we need to find the values of t for which the heights of the baseball and the paintball are the same.

First, let's set the two height equations equal to each other:

-5t^2 + 20t + 15 = 3t + 3

Now, we can solve this quadratic equation for t. To make it easier, let's rearrange it in standard form by bringing all the terms to one side:

-5t^2 + 20t + 15 - (3t + 3) = 0

Simplifying, we get: -5t^2 + 17t + 12 = 0

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's solve it using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = -5, b = 17, and c = 12. Substituting these values into the formula, we get:

t = (-17 ± √(17^2 - 4(-5)(12))) / 2(-5)

Simplifying further, we have:

t = (-17 ± √(289 + 240)) / -10
t = (-17 ± √529) / -10
t = (-17 ± 23) / -10

Now, we have two possible solutions for t:

1. t = (-17 + 23) / -10 = 6 / -10 = -0.6
2. t = (-17 - 23) / -10 = -40 / -10 = 4

Since time cannot be negative in this context, we discard the first solution (t = -0.6).

Therefore, the only valid solution is t = 4 seconds.

To find the height at this time, we substitute t = 4 into one of the height equations (either h = -5t^2 + 20t + 15 or h = 3t + 3).

Using the equation h = -5t^2 + 20t + 15:

h = -5(4)^2 + 20(4) + 15
h = -5(16) + 80 + 15
h = -80 + 80 + 15
h = 15 meters

Therefore, the paintball will hit the baseball when they are both at a height of 15 meters above the ground, and it will happen at t = 4 seconds.