Mathematics, 24.11.2020 07:30, shukriabdisabrie

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5 $t^{2}$ + 14t+2 gives the height h, in meter, of the ball t seconds after it is launched. a)Determine if the ball reaches a height of 14 meters. Explain your answer.

b)How long is the ball in the air?

Answers: 3

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answered: dulce129

Part A)

Part B)

About 2.9362 seconds.

Step-by-step explanation:

The equation $\displaystyle h(t)=-5t^2+14t+2$ models the height h in meters of the ball t seconds after its launch.

Part A)

To determine whether or not the ball reaches a height of 14 meters, we can find the vertex of our function.

Remember that the vertex marks the maximum value of the quadratic (since our quadratic curves down).

If our vertex is greater than 14, then, at some time t, the ball will definitely reach a height of 14 meters.

However, if our vertex is less than 14, then the ball doesn’t reach a height of 14 meters since it can’t go higher than the vertex.

So, let’s find our vertex. The formula for vertex is given by:

$\displaystyle (-\frac{b}{2a},h(-\frac{b}{2a}))$

Our quadratic is:

$\displaystyle h(t)=-5t^2+14t+2$

Hence: a=-5, b=14, and c=2.

Therefore, the x-coordinate of our vertex is:

$\displaystyle x=-\frac{14}{2(-5)}=\frac{14}{10}=\frac{7}{5}$

To find the y-coordinate and the maximum height, we will substitute this value back in for x and evaluate. Hence:

$\displaystyle h(\frac{7}{5})=-5(\frac{7}{5})^2+14(\frac{7}{5})+2$

Evaluate:

$\displaystyle \begin{aligned} h(\frac{7}{2})&=-5(\frac{49}{25})+\frac{98}{5}+2 \\ &=\frac{-245}{25}+\frac{98}{5}+2\\ &=\frac{-245}{25}+\frac{490}{25}+\frac{50}{25}\\&=\frac{-245+490+50}{25}\\&=\frac{295}{25}=\frac{59}{5}=11.8\end{aligned}$

So, our maximum value is 11.8 meters.

Therefore, the ball doesn’t reach a height of 14 meters.

Part B)

To find out how long the ball is in the air, we can simply solve for our t when h=0.

When the ball stops being in the air, this will be the point at which it is at the ground. So, h=0. Therefore:

0=-5t^2+14t+2

A quick check of factors will reveal that is it not factorable. Hence, we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Again, a=-5, b=14, and c=2. Substitute appropriately:

$\displaystyle x=\frac{-(14)\pm\sqrt{(14)^2-4(-5)(2)}}{2(-5)}$

Evaluate:

$\displaystyle x=\frac{-14\pm\sqrt{236}}{-10}$

We can factor the square root:

$\sqrt{236}=\sqrt{4}\cdot\sqrt{59}=2\sqrt{59}$

Hence:

$\displaystyle x=\frac{-14\pm2\sqrt{59}}{-10}$

Divide everything by -2:

$\displaystyle x=\frac{7\pm\sqrt{59}}{5}$

Hence, our two solutions are:

$\displaystyle x=\frac{7+\sqrt{59}}{5}\approx2.9362\text{ or } x=\frac{7-\sqrt{59}}{5}\approx-0.1362$

Since our variable indicates time, we can reject the negative solution since time cannot be negative.

Hence, our zero is approximately 2.9362.

Therefore, the ball is in the air for approximately 2.9362 seconds.

answered: gabbystar517

a) The ball does not reach a height of 14 meters.

b) The ball is in the air for $\frac{7}{5}+\frac{\sqrt{59}}{5}$ , or 2.94 seconds.

Step-by-step explanation:

Part A

We are given the quadratic equation ${h(t)=-5t^2 + 14t + 2}$ .

In order to solve Part A, we need to find the vertex of the quadratic equation.

The parent function of a quadratic function is f(x) =ax^2+bx+c

.The x-coordinate of the vertex is found with the formula $\frac{-b}{2a}$ .The y-coordinate of the vertex is found by setting f(x)

equal to 0 and solving for the variable.

We can define our variables first so they can be substituted into the equation. We use the parent function and align it with our given equation to find the values that align with each variable.

Now, we can use the formula $\frac{-b}{2a}$ to solve for the x-coordinate of the vertex.

$\displaystyle{\frac{-(14)}{2(-5)}}\\\\\frac{-14}{-10}\\\\\frac{14}{10}\\\\x=\frac{7}{5}$

Then, we need to complete the square to solve for the y-coordinate.

$\displaystyle{h(t) = -5\big(t^2 - \frac{14}{5}t\big) + 2}\\\\\big(\frac{b}{2}\big)^2 = \big(\frac{14}{5} \times \frac{1}{2}\big)^2 = \frac{49}{25}\\\\h(t) + \big(\frac{49}{5}\big)\big(-5\big) = -5\big(t^2 - \frac{14}{5}t + (\frac{49}{5}\big)\big) + 2\\\\h(t) - \frac{49}{5} = -5\big(t^2 - \frac{14}{5}t + \frac{49}{25}\big) + 2\\\\h(t) = -5(t^2 - \frac{14}{5}t + \frac{49}{25}) + 2 + \frac{49}{5}\\\\h(t) = -5\big(t - \frac{7}{5}\big)^2 + \frac{59}{5}$

The equation we just solved by completing the square gives us the equation in vertex form - a(x-h)^2+k , where (h, k) is the vertex. Therefore, our x-coordinate has already been solved as $\frac{7}{2}$ and we have determined that our y-coordinate is $\frac{59}{5}.$ This can be converted to a decimal to get 11.8, which we can see is less than 14. Therefore, the ball does not reach a height of 14 meters.

Part B

With us solving for the vertex of the quadratic equation, we now need to solve for approximately how many seconds the ball is in the air. Given the x-coordinate of the vertex, we can plug this value in for the x-variable in the quadratic equation (in this case, the variable is t instead).

I will solve this by completing the square. The other answer shows how to solve this by using the quadratic formula.

$\displaystyle{h(t)=-5t^2+14t+2\\\\0=-5t^2+14t+2}\\\\-5t^2+14t+2=0\\\\t^2-\frac{14}{5}t-\frac{2}{5}=0\\\\t^2-\frac{14}{5}=\frac{2}{5}\\\\t^2-\frac{14}{5}t+\frac{49}{25}=\frac{2}{5}+\frac{49}{25}\\\\\big(t - \frac{7}{5}\big)^2=\frac{2}{5}+\frac{49}{25}\\\\\sqrt{\big(t - \frac{7}{5}\big)^2} = \sqrt{\frac{2}{5}+\frac{49}{25}}\\\\t-\frac{7}{5}=\pm\frac{\sqrt{59}}{5}\\\\t = \frac{7}{5}\pm\frac{\sqrt{59}}{5}$

When simplifying $\frac{7}{5}\pm\frac{\sqrt{59}}{5},$ we can roughly estimate the value of $\sqrt{59}$ by checking perfect squares. A list of the first ten is supplied.

$2^2=4\\$

Because square roots and squaring a value work inversely, we can see if we take the square root of 64, for example, that we will get the value of 8.

So, with this information, we can estimate the value of which 59 will fall between and see if we need to further evaluate the square root or we can safely evaluate the function and see if we get a negative or a positive value.

By using the values above, the value of the square root falls roughly between 7 and 8 when evaluated.

Therefore, we know if we subtract 7 or 8 from $\frac{7}{5}$ , we cannot have a value greater than or equal to 7 as the numerator or we will get a negative value, and time does not exist in negatives. Therefore, we know we must add, not subtract.

Therefore, the ball is in the air for $\frac{7}{5}+\frac{\sqrt{59}}{5}$ seconds, which roughly approximates to about 2.94 seconds.

answered: Guest

A’b’c’ are the coordinates to the new location of the transformed triangle abc. find the distance between the coordinates points to use for your calculations.
Triangle abc is transformed to obtain triangle a’b’c’