THe function h(t)=-16t^2+v0t+h0 describes the height in feet above the ground h(t) of an object thrown vertically from a height of h0 feet, with an initial velocity of v0 feet per second, if there is no air friction and t is the time in seconds sine the object was thrown. A ball is thrown upward from a 100-foot tower at a velocity of 60 feet per second. How many seconds will it take for the ball to reach the ground?

0 = 100 + 60 t - 16 t^2

16 t^2 - 60 t - 100 = 0

4 t^2 - 15 t -25 = 0

(4 t + 5)(t - 5) = 0

t = 5

t=5

To find the time it takes for the ball to reach the ground, we need to determine when the height of the ball, h(t), equals zero.

Given:
h(t) = -16t^2 + v0t + h0

We know the initial height of the ball, h0, is the height of the tower, which is 100 feet. And the initial velocity, v0, is 60 feet per second.

Substituting the given values into the equation, we have:
h(t) = -16t^2 + 60t + 100

To find when the ball reaches the ground, we set h(t) equal to zero and solve for t:
0 = -16t^2 + 60t + 100

This is a quadratic equation. We can solve it by factoring or by using the quadratic formula. Let's use the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

In our equation, a = -16, b = 60, and c = 100. Substituting these values into the equation, we get:

t = (-(60) ± sqrt((60)^2 - 4(-16)(100))) / (2(-16))

Simplifying further:
t = (-60 ± sqrt(3600 + 6400)) / (-32)
t = (-60 ± sqrt(10000)) / (-32)

Since we're interested in the time it takes for the ball to reach the ground, we're only concerned with the positive value of t. Taking the positive solution, we have:

t = (-60 + sqrt(10000)) / (-32)
t = (-60 + 100) / (-32)
t = 40 / (-32)
t = -1.25

Since time cannot be negative, we conclude that it will take the ball 1.25 seconds to reach the ground.

To find the number of seconds it will take for the ball to reach the ground, we need to determine the time when the height, h(t), is zero.

We are given the equation for the height function as:
h(t) = -16t^2 + v0t + h0

In this case, the initial height, h0, is 100 feet, the initial velocity, v0, is 60 feet per second, and we want to find the time, t, when the height, h(t), is zero.

Setting h(t) equal to zero, we can write the equation as:
0 = -16t^2 + 60t + 100

Now, let's solve the quadratic equation to find the values of t:

We can rearrange the equation to standard quadratic form:
16t^2 - 60t - 100 = 0

This quadratic equation can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 16, b = -60, and c = -100.

Plugging in these values and solving, we get:
t = (-(-60) ± √((-60)^2 - 4(16)(-100))) / (2(16))
t = (60 ± √(3600 + 6400)) / 32
t = (60 ± √10000) / 32

Now, let's simplify further:
t = (60 ± 100) / 32

We have two solutions:
t1 = (60 + 100) / 32 = 160 / 32 = 5
t2 = (60 - 100) / 32 = -40 / 32 = -1.25

Since time cannot be negative in this context, we discard the negative value. Therefore, it will take the ball approximately 5 seconds to reach the ground.