If a certain cannon is fired from a height of

7.8 meters above the​ ground, at a certain​ angle, the height of the cannonball above the​ ground, h, in​ meters, at​ time, t, in​ seconds, is found by the function

h(t)=−4.9t2+24.5t+7.8.

What is the maximum height of the cannonball in meters, when is the cannonball at maximum height at in seconds, and when will the cannonball strike the ground at in seconds?

To find the maximum height of the cannonball, we need to determine the vertex of the quadratic equation h(t) = -4.9t^2 + 24.5t + 7.8.

The formula for the x-coordinate of the vertex of a quadratic equation in the form f(t) = at^2 + bt + c is given by x = -b/2a.

In this case, a = -4.9 and b = 24.5.

Using the formula, we can find the x-coordinate of the vertex:

x = -24.5 / (2 * -4.9) = -24.5 / -9.8 = 2.5 seconds (approximately)

Now, substitute this value back into the equation h(t) to find the maximum height:

h(2.5) = -4.9(2.5)^2 + 24.5(2.5) + 7.8 = -30.625 + 61.25 + 7.8 = 38.425 meters (approximately)

Therefore, the maximum height of the cannonball is approximately 38.425 meters.

To find when the cannonball is at maximum height, we know it occurs at t = 2.5 seconds.

Next, to find when the cannonball will strike the ground, we need to solve the equation h(t) = 0:

-4.9t^2 + 24.5t + 7.8 = 0

You can solve this quadratic equation using various methods, such as factoring or using the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

a = -4.9, b = 24.5, c = 7.8

t = (-24.5 ± √(24.5^2 - 4(-4.9)(7.8))) / (2(-4.9))

t = (-24.5 ± √(600.25 + 152.52)) / (-9.8)

t = (-24.5 ± √(752.77)) / (-9.8)

Using a calculator, we can find the solutions are:

t = (-24.5 + √(752.77)) / (-9.8) ≈ 6.42 seconds

t = (-24.5 - √(752.77)) / (-9.8) ≈ 0.64 seconds

Therefore, the cannonball will strike the ground approximately 0.64 seconds and 6.42 seconds after being fired.

To find the maximum height of the cannonball, we need to determine the vertex of the quadratic function h(t) = -4.9t^2 + 24.5t + 7.8. The vertex corresponds to the highest point on the trajectory of the cannonball.

The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (h, k), where h = -b/2a and k = f(h). In our case, a = -4.9, b = 24.5, and c = 7.8.

First, let's find the time when the cannonball reaches its maximum height. The formula for the x-coordinate of the vertex is h = -b/2a, so substituting the values we have:

h = -24.5 / (2 * (-4.9)) = 2.5 s

Therefore, the cannonball reaches its maximum height after 2.5 seconds.

To find the maximum height, we substitute this value of t into the function h(t):

h(2.5) = -4.9(2.5)^2 + 24.5(2.5) + 7.8
= -30.625 + 61.25 + 7.8
= 38.425

Hence, the maximum height of the cannonball is 38.425 meters.

Finally, to determine when the cannonball will strike the ground, we need to find the time at which h(t) = 0. This represents the moment when the cannonball hits the ground.

Setting h(t) = 0, we have:

-4.9t^2 + 24.5t + 7.8 = 0

This is a quadratic equation that can be solved using the quadratic formula. Solving for t, we get two solutions:

t = (-24.5 ± √(24.5^2 - 4 * (-4.9) * 7.8)) / (2 * (-4.9))

Calculating the values under the square root, we find:

t = (-24.5 ± √(600.25 + 152.64)) / -9.8
t = (-24.5 ± √(752.89)) / -9.8
t ≈ (-24.5 ± 27.47) / -9.8

So the two possible times are:

t ≈ (2.97 / -9.8) ≈ -0.303 s
t ≈ (51.97 / -9.8) ≈ -5.306 s

Since time cannot be negative in this context, the cannonball will strike the ground after approximately 5.306 seconds.

In summary, the maximum height of the cannonball is 38.425 meters, it reaches its maximum height after 2.5 seconds, and it will strike the ground after approximately 5.306 seconds.

You know how to find the vertex of a parabola. That will give the maximum height.

It hits the ground when h=0. Sol solve that for t.