Quadratic Projectile Motion Question: height of football given by h = −5t^2 + 20t +2

[cCrane]

First day of the unit, and they assign me homework and I have absolutely no clue where to start. I would assume that I can't just post my question and get an answer which is fine, I would honestly just like some help. I'm in Grade 11 btw and I'm doing quadratics atm.

A quarterback throws a football. The height, h metres, of the ball is given by the equation h = −5t2 + 20t +2 where t is the time in seconds after the ball is thrown.
c) What is the height of the ball 1 second after it is thrown?
d) What is the maximum height of the ball?
e) How long does it take for the ball to reach the maximum height?
f) For how long is the ball more than 10 m above the ground?

 

[Otis]

Hello. Did you already do parts (a) and (b)?

Where did you get stuck on part (c)? They want you to evaluate h, when t=1.

You can google keywords like projectile motion quadratic equation, to find lecture videos and written examples, for your review. If you have trouble finding good lessons or you see something that you don't understand, let us know. Otherwise, please show whatever work you can, or list the specific words you don't understand in the exercise.

This is a tutoring forum. Please read the forum's submissions guidelines; you can start here. Thank you!

PS: We use the caret symbol ^ to show exponents, when typing math expressions. Like this:

h = -5t^2 + 20t + 2

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[cCrane]

Hello. Did you already do parts (a) and (b)?

Where did you get stuck on part (c)? They want you to evaluate h, when t=1.

You can google keywords like projectile motion quadratic equation, to find lecture videos and written examples, for your review. If you have trouble finding good lessons or you see something that you don't understand, let us know. Otherwise, please show whatever work you can, or explain the specific words you don't understand in the exercise.

This is a tutoring forum. Please read the forum's submissions guidelines; you can start here. Thank you!

?

Already answered a and b yeah, I've already watched some videos on Khan academy and one on youtube but I still don't understand. I would assume since h would be meters I woul set h to 0 and maybe flip the equation?

 

[cCrane]

Hello. Did you already do parts (a) and (b)?

Where did you get stuck on part (c)? They want you to evaluate h, when t=1.

You can google keywords like projectile motion quadratic equation, to find lecture videos and written examples, for your review. If you have trouble finding good lessons or you see something that you don't understand, let us know. Otherwise, please show whatever work you can, or explain the specific words you don't understand in the exercise.

This is a tutoring forum. Please read the forum's submissions guidelines; you can start here. Thank you!

PS: We use the caret symbol ^ to show exponents, when typing math expressions. Like this:

h = -5t^2 + 20t + 2

?

Also, I didn't really get stuck I literally just don't know what to do lol

 

[Otis]

… I would assume since h would be meters I would set h to 0 and maybe flip the equation?

Hello again. I'm not sure what you're thinking, when you say, "flip the equation". In which part of the exercise did you think about setting h=0?

Also, I didn't really get stuck I literally just don't know what to do

Before you edited it out, you had mentioned in your first post that you haven't worked quadratic-equation exercises since last year. That's why I suggested review.

Let's start with part (c).

h = -5t^2 + 20t + 2

This is a formula for finding the football's height (that is, the value of h), when you know a value of t (that is, when you know how many seconds have passed since the ball was thrown).

Part (c) tells you how many seconds have passed, since the ball was thrown: t=1 sec

Using that given value of t, we evaluate h by substitution. Here are the steps:

(1) In the formula for h, replace all symbols t with the number 1 (that's the substitution bit)

(2) Do the arithmetic, following the Order of Operations (that's the evaluation bit).

Please show us your work, so we can see what you get for h, when t=1.

?

 

[cCrane]

Hello again. I'm not sure what you're thinking, when you say, "flip the equation". For which part of the exercise, would you think to set h=0?


Before you edited it out, you had mentioned in your first post that you haven't worked quadratic-equation exercises in a while. That's why I suggested review.

Let's start with part (c).

h = -5t^2 + 20t + 2

This is a formula for finding the football's height (that is, the value of h), when you know a value of t (that is, when you know how many seconds have passed since the ball was thrown).

Part (c) tells you how many seconds have passed, since the ball was thrown: t=1 sec

Using that given value of t, we evaluate h by substitution. Here are the steps:

(1) In the formula for h, replace all symbols t with the number 1 (that's the substitution bit)

(2) Do the arithmetic, following the Order of Operations (that's the evaluation bit).

Please show us your work, so we can see what you get for h, when t=1.

?

Okay cool, so the new equation would be: h=−5(1^2)+(20)(1)+2 ? Making the answer 17m?

 

[cCrane]

Assuming I did c) correctly, I am still lost on d) and e), although for f) could I substitute the value 10m for h and then solve the equation?

 

[Otis]

… h=−5(1^2) + (20)(1) + 2 … Making the answer 17m?

You got it! One second after the ball left the quarterback's hand, the ball was 17m above the ground.

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Next: Part (d) asks how high the football goes. If you remember graphing quadratic equations, then you know that the shape of the graph is a parabola (the parabola opens downward, in this exercise) and the highest point is called the vertex. If we knew how many seconds it took for the ball to reach its highest point, then we could find that height using the same method as part (c) -- we would substitute the value of t and evaluate h.

There is a formula for finding the t-coordinate of the vertex point. Do you remember the standard form of a symbolic quadratic polynomial?

at^2 + bt + c, where symbols a,b,c are called coefficients.

The t-coordinate of the vertex point is given by -b/(2a).

Can you identify the value of coefficients a and b, in this exercise? Substitute them into the expression -b/(2a) and do the arithmetic (evaluate). The result is the number of seconds (t) it takes for the ball to reach the vertex. Substitute that value of t into the formula for h and evaluate.

When you're done, you'll have the answer to both part (d) and part (e).

Your turn …

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[Otis]

… for (f) could I substitute the value 10m for h and then solve the equation?

Yes, that's how I would do it!

You will get two values for t. The first value is the number of seconds it takes to reach 10m going up, and the second value is the number of seconds it takes to reach 10m coming down. The difference between those two values of t gives the interval of time when the ball was 10m or more above the ground.

Does your class use graphing calculators? Do you know whether you're supposed to report exact answers versus decimal approximations?

?

 

[cCrane]

You got it! One second after the ball left the quarterback's hand, the ball was 17m above the ground.

---

Next: Part (d) asks how high the football goes. If you remember graphing quadratic equations, then you know that the shape of the graph is a parabola (the parabola opens downward, in this exercise) and the highest point is called the vertex. If we knew how many seconds it took for the ball to reach its highest point, then we could find that height using the same method as part (c) -- we would substitute the value of t and evaluate h.

There is a formula for finding the t-coordinate of the vertex point. Do you remember the standard form of a symbolic quadratic polynomial?

at^2 + bt + c, where symbols a,b,c are called coefficients.

The t-coordinate of the vertex point is given by -b/(2a).

Can you identify the value of coefficients a and b, in this exercise? Substitute them into the expression -b/(2a) and do the arithmetic (evaluate). The result is the number of seconds (t) it takes for the ball to reach the vertex. Substitute that value of t into the formula for h and evaluate.

When you're done, you'll have the answer to both part (d) and part (e).

Your turn …

?

Okay So when I convert this equation into vertex form I'm getting the points -2 and -18, which doesn't make any sense because they are negative.

 

[cCrane]

Yes, that's how I would do it!

You will get two values for t. The first value is the number of seconds it takes to reach 10m going up, and the second value is the number of seconds it takes to reach 10m coming down. The difference between those two values of t gives the interval of time when the ball was 10m or more above the ground.

Does your class use graphing calculators? Do you know whether you're supposed to report exact answers versus decimal approximations?

?

No, the simple "17m" is perfectly fine. And for f this is what I did: I subbed 10 for h making the equation 10=5t^2+20t+2 , I put it into my calculator and it gave me two different equations? One being; t=2(5-sqrt15) / 5 and t=2(5+sqrt15)/5 ...? I calculated the positive one(2nd one) and it gave me around 3.5 for t... Pretty sure I'm doing something wrong lol.

 

[Otis]

… for (f) this is what I did: I subbed 10 for h making the equation 10=5t^2+20t+2 , I put it into my calculator and it gave me two different [solutions for t]?

Yes, that's correct. There are two times, when the ball is 10m above the ground. Please read post #9, again.

… t=2(5-sqrt15)/5 and t=2(5+sqrt15)/5 ... I calculated the positive one and it gave me around 3.5 for t …

Each of those expressions for t is a positive number, and you need both of them to answer part (f).

Since you're using decimal approximations, I would round intermediate results to three decimal places. I would round the final answer to one decimal place, unless they've given you different rounding instructions and you forgot to post them.

EDIT: Removed question about finishing parts (d) and (e)

?

 

[Otis]

Okay So when I convert this equation into vertex form I'm getting the points -2 and -18, which doesn't make any sense because they are negative.

I didn't see post #10, prior to my last reply, because it was pending approval.

You're talking about substituting the values of b and a into the expression -b/(2a), yes? You should not get two results because -b/(2a) is only one expression. Instead of describing results, please type out your work. You're not using proper terminology, so I can't be sure what you've actually done unless you show it.

What are the values of a,b,c in this exercise?

How did you get -b/(2a) = -2 ?

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