Question

Antoine stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), x seconds after Antoine threw it, is modeled

by: h (x) = -2x7 + 4x + 16
a) How many seconds after being thrown will the ball hit the ground?
b) At what time did the ball reach its maximum height?
c) what is the maximum height the ball reached.
d) What is the height of the ball from where Antoine throws the ball?

Answer 1

Explanation:

a) The ball hits the ground when its height, h(x), is equal to 0. We can set the equation h(x) = 0 and solve for x:

-2x^2 + 4x + 16 = 0

Dividing by -2:

x^2 - 2x - 8 = 0

Using the quadratic formula:

x = (2 ± √(2^2 - 4(1)(-8))) / 2

x = (2 ± √(36)) / 2

x = 2 ± 3

Therefore, the ball will hit the ground 5 seconds after being thrown (using the positive value of x).

b) The maximum height of the ball occurs at the vertex of the parabolic function. We can find the x-coordinate of the vertex using the formula:

x = -b / (2a)

where a = -2 and b = 4. Substituting these values, we get:

x = -4 / (2(-2)) = 1

Therefore, the ball reaches its maximum height 1 second after being thrown.

c) To find the maximum height, we can substitute the value of x = 1 into the equation for h(x):

h(1) = -2(1)^2 + 4(1) + 16

h(1) = 18

Therefore, the maximum height of the ball is 18 meters.

d) The initial height of the ball is the value of h(x) when x = 0. Substituting x = 0 into the equation for h(x), we get:

h(0) = -2(0)^2 + 4(0) + 16

h(0) = 16

Therefore, the height of the ball from where Antoine throws the ball is 16 meters.