A ball is thrown straight up from a bridge over a river and falls into the water. The height, h, in meters, of the ball above the water t seconds after being thrown is approximately modeled by the relation h=-5t^2+10t+35.

Can you please help me !

Show your work.

a) What is the maximum height of the ball above the water? (1 mark)

b) How long does it take for the ball to reach the maximum height? (1 mark)
See below for what I need for the answer to be valid

c) After how many seconds does the ball hit the water? (1 mark)
I need help with this question with an actual explanation of you got the answer as I can't seem to figure it out.

d) How high is the bridge above the river? (1 mark)

The height of the bridge above the water would be 35 meters

h=-5t^2+10t+35
h=-5(0)^2+10(0)+35
h=0+0+35
h=35

(b) t = -b/2a = 10/10 = 1

(a) find h(1)
(c) solve for t when h=0
-5t^2+10t+35 = 0
or,
t^2-2t-7 = 0
t = (2±√30)/2 (use quadratic formula)
(d) correct

oops. That's

(2±√32)/2 = 1±√8

a) To find the maximum height of the ball, we need to determine the vertex of the quadratic equation. The vertex of a quadratic equation in the form h=-5t^2+10t+35 can be found using the formula t = -b/(2a), where a = -5 and b = 10.

Using this formula, we have t = -10/(2 * -5) = -10/-10 = 1.

Now substitute t = 1 into the equation to find the maximum height:

h = -5(1)^2 + 10(1) + 35
h = -5 + 10 + 35
h = 40

Therefore, the maximum height of the ball above the water is 40 meters.

b) The time it takes for the ball to reach the maximum height can be determined from the vertex, which we found to be t = 1 second.

Therefore, it takes 1 second for the ball to reach the maximum height.

c) To find out when the ball hits the water, we need to set h = 0 in the height equation h = -5t^2 + 10t + 35 and solve for t.

0 = -5t^2 + 10t + 35

Rearranging the equation, we get:

5t^2 - 10t - 35 = 0

This is a quadratic equation that we can solve using factoring, completing the square, or the quadratic formula. In this case, we'll solve it using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a).

Plugging in the values, we have:

t = (-(-10) ± √((-10)^2 - 4 * 5 * -35)) / (2 * 5)
t = (10 ± √(100 + 700)) / 10
t = (10 ± √(800)) / 10
t = (10 ± 28.28) / 10

Simplifying, we have two solutions:

t = (10 + 28.28) / 10 = 3.83
t = (10 - 28.28) / 10 = -1.83

Since time cannot be negative in this context, we discard the negative value.

Therefore, the ball hits the water approximately 3.83 seconds after it was thrown.

d) The height of the bridge above the water can be determined by looking at the initial equation h = -5t^2 + 10t + 35, where t = 0. Substituting t = 0 into the equation, we have:

h = -5(0)^2 + 10(0) + 35
h = -5(0) + 0 + 35
h = 35

Therefore, the height of the bridge above the river is 35 meters.

Sure, let's go through each part step by step.

a) To find the maximum height of the ball above the water, we need to determine the vertex of the parabolic function. The formula for finding the x-coordinate of the vertex is given by the equation: x = -b / (2a), where a, b, and c are the coefficients in the quadratic equation.

In this case, our equation is h = -5t^2 + 10t + 35. By comparing this to the standard form of a quadratic equation (h = at^2 + bt + c), we can identify that a = -5 and b = 10.

Using the formula, we find: t = -10 / (2 * -5) = -10 / -10 = 1.

So, the maximum height of the ball above the water occurs after 1 second.

To determine the height, we substitute the value of t back into the equation: h = -5(1)^2 + 10(1) + 35 = -5 + 10 + 35 = 40 meters.

Therefore, the maximum height of the ball above the water is 40 meters.

b) To find how long it takes for the ball to reach the maximum height, we have already determined that the ball reaches its maximum height after 1 second.

Therefore, it takes 1 second for the ball to reach the maximum height.

c) To determine when the ball hits the water, we need to find the time when the height, h, is equal to 0. We set the equation h = -5t^2 + 10t + 35 equal to 0.

-5t^2 + 10t + 35 = 0

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. I will show you how to solve it using the quadratic formula.

The quadratic formula is given by: t = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = -5, b = 10, and c = 35. Plugging these values into the quadratic formula gives us:

t = (-10 ± √(10^2 - 4(-5)(35))) / (2 * -5)

t = (-10 ± √(100 + 700)) / (-10)

t = (-10 ± √(800)) / (-10)

t = (-10 ± 28.28) / (-10)

Solving for t gives us two possible values:

t ≈ (-10 + 28.28) / (-10) ≈ -1.73

t ≈ (-10 - 28.28) / (-10) ≈ 3.83

Since time cannot be negative in this context, we discard the negative value.

Therefore, the ball hits the water approximately 3.83 seconds after it is thrown.

d) The height of the bridge above the water can be found by substituting t = 0 into the equation h = -5t^2 + 10t + 35.

h = -5(0)^2 + 10(0) + 35

h = 0 + 0 + 35

h = 35

Therefore, the height of the bridge above the river is 35 meters.