A toy cannon ball is launched from a cannon on, №9884885, 03.09.2021 23:06

Mathematics

A toy cannon ball is launched from a cannon on top of a platform. the equation h(t)=-5t^2+15t+10 gives the height h, in meters, of the ball t seconds after it is launched. A) Does the ball reach a height of 24m B)Explain your answer ASAP

Step-by-step answer

P Answered by PhD

3

Hello,

This is a parabola.

The vertex is for x= 15/2*5=3/2, and then the maximum is reached for

$h(3/2)=-5(9/4)+15(3/2)+10=\dfrac{-45+90+40}{4}=\dfrac{85}{4}$

As 85/4 < 24 the ball does not reach a height of 24 m

thanks

Mathematics

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5 + 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?

Step-by-step answer

P Answered by PhD

11

Part A)

No

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.

Mathematics

A toy cannon ball is launched from a cannon on top of a platform. the equation h(t)=-5t^2+15t+10 gives the height h, in meters, of the ball t seconds after it is launched. A) Does the ball reach a height of 24m B)Explain your answer ASAP

Step-by-step answer

P Answered by PhD

3

Hello,

This is a parabola.

The vertex is for x= 15/2*5=3/2, and then the maximum is reached for

$h(3/2)=-5(9/4)+15(3/2)+10=\dfrac{-45+90+40}{4}=\dfrac{85}{4}$

As 85/4 < 24 the ball does not reach a height of 24 m

thanks

Mathematics

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5 + 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?

Step-by-step answer

P Answered by PhD

11

Part A)

No

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.

Mathematics

Atoy cannon ball is launched from a cannon on top of a platform. the equation h(t) = -5t^2 + 20t + 4 gives the height h, in meters, of the ball t seconds after it is launched. a. what equation can be used to tell whether the ball reaches a height of 12 m ? b. does the ball reach a height of 12m? (yes or no)

Step-by-step answer

P Answered by PhD

15

A) $5t^{2} -20t+8$ =0

B)t= 0.45 seconds or t= 3.5 seconds

If time is greater than 0.45 seconds then ball will reach height 12m and higher.

Step-by-step explanation:

Given Equation:

h(t)= $-5t^{2} +20t+4$ ------------------------------(Equation 1)

a) Equation to tell if ball reaches height of 12m . for that :

h= 12m

put in Equation 1.

12 = $-5t^{2} +20t+4$

or $-5t^{2} +20t+4$ =12

or $-5t^{2} +20t+4$ -12 = 0

$-5t^{2} +20t-8$ =0

or $5t^{2} -20t+8$ = 0 -----------------------------------(Equation 2)

This Equation tells if ball reaches height of 12m

b) Does ball reaches height of 12m :

For that, the value of time can be found out from the equation above,

$5t^{2} -20t+8$ = 0

It can be solved using the quadratic formula:

$t= \frac{(-b) +_{-} \sqrt{b^{2} -4ac } }{2a}$

$t= \frac{(20) +_{-} \sqrt{(-20)^{2} -4(5)(8) } }{2(5)}$

t= 0.45 seconds or t= 3.5 seconds

If time is greater than 0.45 seconds then ball will reach height 12m and higher.

Mathematics

3. REI.b4_Part A. A toy cannon ball is launched on top of a platform. The equation h(t) =-8t +18t +6 gives the height, h in meters, of the ball t seconds after it is launched. What equation can be used to tell whether the ball reaches the height of 10 meters? *|| A. -8t^2 +18t +6 =0 B. -8t^2 +18t +6 =-10 o C.-8t^2 +18t +6 = 12 o D. -8t^2 +18t +6 = 10

Step-by-step answer

P Answered by PhD

The correct option is;

D. -8·t² + 18·t + 6 = 10

Step-by-step explanation:

The given equation of the height toy cannon launched on top of a platform is given as follows;

h(t) = -8·t² + 18·t + 6...(1)

Where "h" represents the height in meters, at time "t" seconds after the ball is launched

When the height is 10 meters, we can write as follows;

The height of the ball, h(t) = 10

Therefore, by substituting the value of h(t) in equation (1), h(t) = -8·t² + 18·t + 6, when h(t) = 10 we get;

-8·t² + 18·t + 6 = 10.

Mathematics

3. REI.b4_Part A. A toy cannon ball is launched on top of a platform. The equation h(t) =-8t +18t +6 gives the height, h in meters, of the ball t seconds after it is launched. What equation can be used to tell whether the ball reaches the height of 10 meters? *|| A. -8t^2 +18t +6 =0 B. -8t^2 +18t +6 =-10 o C.-8t^2 +18t +6 = 12 o D. -8t^2 +18t +6 = 10

Step-by-step answer

P Answered by PhD

The correct option is;

D. -8·t² + 18·t + 6 = 10

Step-by-step explanation:

The given equation of the height toy cannon launched on top of a platform is given as follows;

h(t) = -8·t² + 18·t + 6...(1)

Where "h" represents the height in meters, at time "t" seconds after the ball is launched

When the height is 10 meters, we can write as follows;

The height of the ball, h(t) = 10

Therefore, by substituting the value of h(t) in equation (1), h(t) = -8·t² + 18·t + 6, when h(t) = 10 we get;

-8·t² + 18·t + 6 = 10.

Mathematics

You can buy jars of the same jam in 2 sizes large jar:440g for £1.54 small jar:340g for £1.26 By considering the price per gram, work out which jar is the best value for money. Working must be shown

Step-by-step answer

P Answered by PhD

Answer: 440 grams for 1.54 is the better value
Explanation:
Take the price and divide by the number of grams
1.54 / 440 =0.0035 per gram
1.26 / 340 =0.003705882 per gram
0.0035 per gram < 0.003705882 per gram

Mathematics

A new school randomly assigns a six-character student code to every prospective student. The school uses the characters 1, 2, 3, 4, 5, X, Y, and Z. If none of the student codes contain a repeated character, what is the total number of codes that can be randomly assigned?

Step-by-step answer

P Answered by PhD

The total nom of code that can be used is equal to 5+3 = 8

Mathematics

The ratio of tables to chairs at the furniture store is 1:6. There are 30 chairs. How many tables are there?

Step-by-step answer

P Answered by PhD

The solution is in the following image

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t)= -5t^2+30t+ 8 gives the height h, in meters.

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