SOLUTION: A ball is thrown upward from the top of a building. The ball's height above the ground after t seconds is given by the function: h(t) = -16t^2 + 48t +32.
a. What is the initial
Algebra ->
Test
-> SOLUTION: A ball is thrown upward from the top of a building. The ball's height above the ground after t seconds is given by the function: h(t) = -16t^2 + 48t +32.
a. What is the initial
Log On
Question 730246: A ball is thrown upward from the top of a building. The ball's height above the ground after t seconds is given by the function: h(t) = -16t^2 + 48t +32.
a. What is the initial height (i.e. the height of the building)?
b. How high did the ball go?
c. When does the ball hit the ground? Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
(a)
The instant the ball is thrown upward is
called . Plug this value into the equation. is the height of the building. ft
(b)
The path of the ball is a parabola. The t-coordinate
of the vertex is at when the form
is like the given equation
Plug this back into equation to find
The max height is 68 ft
(c)
The height is zero when it hits the ground, so
Use the quadratic formula
The ball hits the ground in 3.562 sec
---------------------------------
Here's the plot of the equation:
