A ball is thrown upwards, from a building. It's height(h), in meters above the ground seconds(t) after being thrown is given by h(t)=-5t^2+15t+50

a) How tall is the building? (explain your answer)

b) How long does it take the ball to reach the ground?

c) For how long is the ball 55m above the ground?

You know that the initial height is the constant term. So, the building is 50m tall

The ball hits the ground when the height is zero. So, just solve for t when h=0

I assume you want to know how long the ball is at least 55m above the ground. It is only at 50m for an instant. So, we want

-5t^2+15t+50 >= 55
1/2 (3-√5) <= t <= 1/2 (3+√5)

http://www.wolframalpha.com/input/?i=+-5t^2%2B15t%2B50+%3E%3D+55

a) h(t)=-5t^2+15t+50

h(0)=0+0+50
h=50

a) To find the height of the building, we need to determine the initial height of the ball when it was thrown upwards. In this case, the initial height is given by the constant term in the equation, which is 50 meters. Therefore, the height of the building is 50 meters.

b) To find the time it takes for the ball to reach the ground, we need to determine when the height of the ball is equal to 0. We can set h(t) = 0 and solve for t.

-5t^2 + 15t + 50 = 0

To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac))/(2a)

For this equation, a = -5, b = 15, and c = 50.

t = (-15 ± √(15^2 - 4(-5)(50)))/(2(-5))
= (-15 ± √(225 + 1000))/(-10)
= (-15 ± √1225)/(-10)
= (-15 ± 35)/(-10)

Taking both solutions into account, we have:

t1 = (-15 + 35)/(-10) = 2

t2 = (-15 - 35)/(-10) = 5

Since time cannot be negative, the ball takes 5 seconds to reach the ground.

c) To find how long the ball is 55 meters above the ground, we need to set h(t) = 55 and solve for t.

-5t^2 + 15t + 50 = 55

-5t^2 + 15t - 5 = 0

Dividing by -5 to simplify, we get:

t^2 - 3t + 1 = 0

Using the quadratic formula:

t = (-(-3) ± √((-3)^2 - 4(1)(1)))/(2(1))
= (3 ± √(9 - 4))/2
= (3 ± √5)/2

This gives two possible solutions for t:

t1 = (3 + √5)/2

t2 = (3 - √5)/2

So, the ball is 55 meters above the ground for (3 + √5)/2 seconds and (3 - √5)/2 seconds.

To find the answers to these questions, we will utilize the provided function h(t) = -5t^2 + 15t + 50, which represents the height of the ball above the ground at any given time t.

a) How tall is the building?

To determine the height of the building, we need to find the value of h when t is 0. Plugging t = 0 into the equation gives us:

h(0) = -5(0)^2 + 15(0) + 50
h(0) = 0 + 0 + 50
h(0) = 50

Therefore, the building's height is 50 meters.

b) How long does it take the ball to reach the ground?

The ball hits the ground when its height is h = 0. We need to find the time at which h(t) = 0. To do this, we set the equation equal to zero and solve for t:

0 = -5t^2 + 15t + 50

This is a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values from our equation, we have:

t = (-(15) ± √((15)^2 - 4(-5)(50))) / (2(-5))
t = (-15 ± √(225 + 1000)) / (-10)
t = (-15 ± √(1225)) / (-10)
t = (-15 ± 35) / (-10)

The two solutions are:
t1 = (-15 + 35) / (-10) = 20 / (-10) = -2
t2 = (-15 - 35) / (-10) = -50 / (-10) = 5

Since time cannot be negative in this context, we discard the negative solution. Therefore, the ball takes 5 seconds to reach the ground.

c) For how long is the ball 55m above the ground?

We need to find the time (t) when the height of the ball (h) is 55 meters. We set h(t) = 55 and solve for t:

55 = -5t^2 + 15t + 50

Re-arranging the equation to form a quadratic equation:

5t^2 - 15t + 5 = 0

Solving this equation using factoring, completing the square, or the quadratic formula gives us two solutions. Let's use the quadratic formula again:

t = (-(-15) ± √((-15)^2 - 4(5)(5))) / (2(5))
t = (15 ± √(225 - 100)) / 10
t = (15 ± √(125)) / 10
t = (15 ± √(5^2 * 5)) / 10
t = (15 ± 5√5) / 10

The two solutions are:
t1 = (15 + 5√5) / 10
t2 = (15 - 5√5) / 10

Therefore, the ball is 55 meters above the ground for t1 seconds and t2 seconds, where t1 and t2 are the solutions obtained above.