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.
h = -16t^2 + 48t + 160.
a. V = Vo + g*Tr.
0 = 48 - 32Tr, Tr = 1.5 s. = Rise time.
b. h = -16*1.5^2 + 48*1.5 + 160 = 196 Ft. above gnd.
c. h = 0.5g*Tf^2.
196 = 16Tf^2, Tf^2 = 12.25, Tf = 3.5 s. = Fall time.
Tr + Tf = 1.5 + 3.5 = 5 s. = Time to hit gnd.
d. h = -16*2^2 + 48*2 + 160 =
To find the answers to these questions, we need to work with the given quadratic equation and use some concepts from algebra and calculus. Let's break down each question and explain how to solve them step by step:
a) When does the ball reach its maximum height?
To determine when the ball reaches its maximum height, we need to find the vertex of the quadratic equation. The vertex of a quadratic function in the form f(t) = at^2 + bt + c is (-b/2a, f(-b/2a)).
For our given equation h(t) = -8t^2 + 48t + 160, we can identify a = -8, b = 48, and c = 160. To find the time (t) when the ball reaches its maximum height, we use the formula t = -b/2a.
Substituting the values, we have t = -48/(2*(-8)). Simplifying this expression, t = -48/(-16) = 3 seconds.
Therefore, the ball reaches its maximum height after 3 seconds.
b) What is the ball's maximum height?
To find the maximum height of the ball, we substitute the value of t = 3 into the equation h(t) = -8t^2 + 48t + 160:
h(3) = -8(3)^2 + 48(3) + 160 = -72 + 144 + 160 = 232 feet.
Thus, the ball reaches a maximum height of 232 feet.
c) How long does it take before the ball comes back and hits the ground?
To determine when the ball hits the ground, we need to find the time when the height (h) is equal to 0. We set h(t) = 0 and solve for t.
-8t^2 + 48t + 160 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/2a
Substituting the values a = -8, b = 48, and c = 160 into the equation, we get:
t = (-48 ± √(48^2 - 4*(-8)*160))/2*(-8)
Simplifying this expression will give you two possible values for t. Ignore any negative values since time cannot be negative in this context.
Hence, the time it takes before the ball hits the ground can be determined by the positive value of t obtained from solving the quadratic equation.
d) How high will the ball be at 2 seconds?
To find the height of the ball at t = 2 seconds, we substitute t = 2 into the equation h(t) = -8t^2 + 48t + 160:
h(2) = -8(2)^2 + 48(2) + 160 = -32 + 96 + 160 = 224 feet.
Therefore, at 2 seconds, the ball will be 224 feet high.