If a ball is thrown upward from a building 30 m tall and the ball has a vertical velocity of 25 m/s, then its approximate height above the ground t seconds later is given by h(t) = 30 + 25t - 5t^2

a. After how many seconds does the ball hit the ground?
b. What is the domain of h?
c. How high does the ball go?

a. use the equation given, h(t)=0, so

0=30+25t-5t^2 solve for t.
h domain is zero to max height
max height will be when velocity is zero (at top)
V=25-10t but v is zero, solve for t at max height, then put that t in the original h(t) equation.

a. Well, if the ball hits the ground, it means that its height is 0. So, to find out how many seconds it takes for the ball to hit the ground, we can set h(t) = 0 and solve for t. In other words, we want to find the value of t when the equation 30 + 25t - 5t^2 = 0. But hey, instead of solving it ourselves, let's throw that equation onto the ground and see if it hits around 0 seconds!

b. Now, let's talk about the domain of h. We're dealing with time here, which means we can't go back in time (unless you're a time-traveling clown, of course). So, the domain of h is all the real numbers greater than or equal to 0. Don't waste your time going backward!

c. To find out how high the ball goes, we need to find the maximum height. Mathematically speaking, the maximum height occurs at the vertex of the parabolic function. So, let's not get scared of heights and use a little bit of math to find the value of t at the vertex. The vertex of the parabola h(t) = 30 + 25t - 5t^2 is given by the formula t = -b/2a. In this case, a = -5 and b = 25. Plug those values into the formula and you'll find the t-value at the vertex. But don't worry, there's no need to juggle numbers here.

a. To find the time it takes for the ball to hit the ground, we need to set h(t) equal to zero and solve for t. The equation is:

0 = 30 + 25t - 5t^2

Rearrange the equation to set it equal to zero:

5t^2 - 25t - 30 = 0

Now, we can factor the quadratic equation:

(t - 3)(5t + 10) = 0

Setting each factor equal to zero:

t - 3 = 0 or 5t + 10 = 0

Solving for t in each equation:

t = 3 or 5t = -10

Since time cannot be negative in this context, we discard the negative value. Therefore, the ball hits the ground after 3 seconds.

b. The domain of h is the set of all possible input values of t. In this case, the domain of h is all real numbers, as there is no restriction on the possible values of t.

c. To determine the maximum height the ball reaches, we can use the fact that the vertex of a quadratic function represents its maximum or minimum point.

In the given equation, h(t) = 30 + 25t - 5t^2, the coefficient of t^2 is negative (-5). Therefore, the parabola opens downward, and the vertex represents the maximum point.

The x-coordinate of the vertex can be found using the formula: t = -b / (2a), where a = -5 and b = 25. Plugging in the values:

t = -25 / (2(-5))
t = -25 / (-10)
t = 2.5

To find the height at this time, substitute t = 2.5 back into the equation:

h(2.5) = 30 + 25(2.5) - 5(2.5)^2
h(2.5) = 30 + 62.5 - 5(6.25)
h(2.5) = 30 + 62.5 - 31.25
h(2.5) = 61.25

Therefore, the ball reaches a maximum height of 61.25 meters.

a. To find the time when the ball hits the ground, we need to determine when its height above the ground is zero. We can set h(t) = 0 and solve for t:

0 = 30 + 25t - 5t^2

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

For the equation h(t) = 30 + 25t - 5t^2, the coefficients are:
a = -5
b = 25
c = 30

Substituting these values into the quadratic formula, we have:
t = (-(25) ± √((25)^2 - 4(-5)(30))) / (2(-5))
t = (-25 ± √(625 + 600)) / (-10)
t = (-25 ± √1225) / (-10)
t = (-25 ± 35) / (-10)

This gives two possible solutions:
t1 = (-25 + 35) / (-10) = 1
t2 = (-25 - 35) / (-10) = 6

Since time cannot be negative, we ignore the negative solution. Therefore, the ball hits the ground after approximately 6 seconds.

b. The domain of h represents the possible values of t for which the function h(t) is defined. In this case, the function h(t) is defined for all real numbers since there are no restrictions on t. Therefore, the domain of h is (-∞, ∞), which means any real number can be inputted into the function.

c. To find the maximum height of the ball, we can observe that the equation h(t) is in the form of a quadratic function: h(t) = -5t^2 + 25t + 30.

The maximum height occurs at the vertex of this quadratic function, which is given by the formula t = -b / (2a).

For the equation h(t) = -5t^2 + 25t + 30, the coefficients are:
a = -5
b = 25

Substituting these values into the formula, we have:
t = -25 / (2(-5))
t = -25 / (-10)
t = 2.5

To find the height, we can substitute t = 2.5 into the equation h(t):
h(2.5) = 30 + 25(2.5) - 5(2.5)^2
h(2.5) = 30 + 62.5 - 5(6.25)
h(2.5) = 30 + 62.5 - 31.25
h(2.5) = 61.25

Therefore, the ball reaches a maximum height of approximately 61.25 meters.