Can someone please help me, I've been stuck on this question for a hour

[Teemo]

I'm not sure where to begin and what I should do exactly.

A diver followed a path defined by h(t) = −4.9t² + 3t + 10 in her dive, where t is the time,
in seconds, and h represents her height above the water, in meters.

a) What was the maximum height the diver reached?

b) For how long was the diver in the air?

 

[lev888]

Initially h is 10 (why?). The diver jumps up and is above water for some time. Then she reaches water. You need to find max height and the time when h became 0 again. What are you studying? Functions, derivative?

 

[Deleted member 4993]

I'm not sure where to begin and what I should do exactly.

A diver followed a path defined by h(t) = −4.9t² + 3t + 10 in her dive, where t is the time,
in seconds, and h represents her height above the water, in meters.

a) What was the maximum height the diver reached?

b) For how long was the diver in the air?

If I were to solve your assignment, I would first draw an approximate sketch of the "diver's path" - plotting 'h' vs. 't'.

What is the height of the diving board - calculated from 'h(t)' at t=0? ← h(0) = -4.9*(0)^2 + 3*(0) +10

What is the derivative of h(t) with respect to time? How is that derivative related to the maximum height reached by the diver?

At what value of 't = t₁' you have h(t₁)= 0 .....←..... -4.9*t₁² + 3*t₁ + 10 = 0

continue....

 

[JeffM]

Because you posted this in algebra, you may not know what a derivative is, and you may be studying parabolas. Is that what you are studying?

 

[lex]

Yes, should this be simply about quadratics, to do the first part you could find the height at t=0 (the original height), put the expression for h equal to this value and solve the quadratic equation to find the value of t (when the height equals this original height again).
Halve this value of t in order to get the time when she was at her greatest height (we know it will be half this time, since the quadratic graph of h against t is symmetrical).
Substitute this value of t into the expression for h, to get the maximum height.

Part (b) find the time when h=0.

 

[Teemo]

Because you posted this in algebra, you may not know what a derivative is, and you may be studying parabolas. Is that what you are studying?

yea I think so, sorry about that

 

[Teemo]

If I were to solve your assignment, I would first draw an approximate sketch of the "diver's path" - plotting 'h' vs. 't'.

What is the height of the diving board - calculated from 'h(t)' at t=0? ← h(0) = -4.9*(0)^2 + 3*(0) +10

What is the derivative of h(t) with respect to time? How is that derivative related to the maximum height reached by the diver?

At what value of 't = t₁' you have h(t₁)= 0 .....←..... -4.9*t₁² + 3*t₁ + 10 = 0

continue....

thank you!

 

[JeffM]

yea I think so, sorry about that

So you apply what you have learned about parabolas.

Where is the vertex?

What is the value of the vertex?

Why is the vertex important?

When the diver reaches the water, what is her height above the water?

 

[Steven G]

Initially h is 0 (why?). The diver jumps up and is above water for some time. Then she reaches water. You need to find max height and the time when h became 0 again. What are you studying? Functions, derivative?

lev, I am sorry but while it is true that initially t=0 in this case h does not start at 0. In fact h(0) = 10.

 

[lev888]

lev, I am sorry but while it is true that initially t=0 in this case h does not start at 0. In fact h(0) = 10.

Thanks. I think I was using the phone and +10 got cut off.

 

[Deleted member 4993]

I'm not sure where to begin and what I should do exactly.

A diver followed a path defined by h(t) = −4.9t² + 3t + 10 in her dive, where t is the time,
in seconds, and h represents her height above the water, in meters.

a) What was the maximum height the diver reached?

b) For how long was the diver in the air?

If you do not know calculus - then - along with the approximate sketch, the best numerical method would be to "complete square" .

Are you familiar with that method? You convert

h(t) = −4.9*t² + 3*t + 10

to

h(t) = -c*(t - t_o)² - h₁

Please share your work (calculation of c, t_o and h₁) indicating where you need help.

 

[Steven G]

lev, I am sorry but while it is true that initially t=0 in this case h does not start at 0. In fact h(0) = 10.

Oh, I completely understand now.

 

[Steven G]

You are given a negative parabola that always has a max at t = -b/a. Once you have the t-value for the maximum h value, you need to find the corresponding h-value. That will be the max value for h.
Try to do the 2nd part on your own.

 

[Otis]

… parabola … has a max at t = -b/a …

You meant to type t = -b/(2a).

That's how I would answer part (a), too, Jomo. For a parabola that opens upwards or downwards, the vertex point's x-coordinate (or, t-coordinate, in this thread) is -b/(2a).

To find the y-coordinate, we substitute -b/(2a) for t in at^2+bt+c. That simplifies to c - b^2/(4a).

The paraola's vertex point is located at ( -b/(2a), c - b^2/(4a) )

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