Suppose you throw a ball up in the air from the top of a 160 foot building. It's height h in feet after t seconds is given by the function h= -8t^ + 48t + 160

a) When does the ball reach it's maximum height?

b) What is the ball's maximum height?

c) How long does it take before the ball comes back and hits the ground?

d) How high will the ball be at 2 seconds?

I know it's a quadratic equation but not sure how to set it up to find solutions by factoring or a parabola.

Ignoring the fact that this equation would violate the laws of gravity on earth ....

find the vertex in the form (t,h)
the value of t is your a) answer, and the value of h is your b) answer.
Use the method that was taught to you.

c) set -8t^2 + 48t + 160 = 0
divide by -8
t^2 - 3t - 20 = 0
solve using the quadratic equation, of course you would use only the positive value of t.

d) sub in t = 2 into your equation.

If I divide the quadratic equation by a -8 wouldn't it be

t^ - 6t - 20 = 0

t^2 - 6t - 20 = 0

yes, you are right, my mental error or senior moment.

and that can't be factored

To find the answers to the given questions, we need to understand the properties of the quadratic equation and how it relates to the motion of the ball.

The equation h = -8t^2 + 48t + 160 represents the height of the ball (h) as a function of time (t).

a) To determine when the ball reaches its maximum height, we need to find the vertex of the parabola. The vertex is given by the equation t = -b/2a.

In this case, a = -8 and b = 48. Substituting these values into the equation, we have t = -48/(2*(-8)). Simplifying further, t = 3.

Therefore, the ball reaches its maximum height after 3 seconds.

b) To find the ball's maximum height, we substitute the value of t we found (t = 3) back into the equation h = -8t^2 + 48t + 160.

Plugging in t = 3, we get h = -8(3)^2 + 48(3) + 160. Evaluating this expression, h = 88.

Therefore, the ball's maximum height is 88 feet.

c) To determine how long it takes for the ball to hit the ground, we need to find the value of t when h equals zero.

Setting h = 0 in the equation -8t^2 + 48t + 160 = 0, we can solve for t using either factoring, completing the square, or the quadratic formula.

Using the quadratic formula, t = (-b ± √(b^2 - 4ac))/(2a), we substitute a = -8, b = 48, and c = 160 into the equation.

Applying the quadratic formula, t = (-48 ± √(48^2 - 4*(-8)*160))/(2*(-8)). Simplifying this expression, t = (-48 ± √(2304 + 5120))/(-16).

Further simplification gives t = (-48 ± √7424)/(-16). Since √7424 is approximately 86.15, we have t ≈ (-48 ± 86.15)/(-16).

By evaluating both possibilities, t ≈ (-48 + 86.15)/(-16) or t ≈ (-48 - 86.15)/(-16), we find that t ≈ 8.86 or t ≈ -2.14.

The negative value of time (-2.14) does not make sense given the context of the problem, so we discard it.

Hence, the ball takes approximately 8.86 seconds to hit the ground.

d) To determine the height of the ball at 2 seconds, we substitute t = 2 into the equation h = -8t^2 + 48t + 160.

Substituting t = 2, we have h = -8(2)^2 + 48(2) + 160. Simplifying, h = 112.

Therefore, the ball will be at a height of 112 feet after 2 seconds.

By following these steps, you can find the solutions to all the given questions related to the ball's motion.