What is the Velocity of a Ball Thrown Vertically Upward at a Height of 96 Feet?

[morphine]

Homework Statement

If a ball is thrown vertically upward with a velocity of 80 ft/sec,
then its height after t seconds is given by
s = 80t - 16t²
. What is the velocity of the ball
when it is 96 feet above the ground on its way up?

Homework Equations

The Attempt at a Solution

I have found my way to s'(t) = -16h -32t + 80
which I believe to be right. But now I'm not sure what to do. Do I need to find the second derivative?

 

[Phyisab****]

Where does 16h come from in your equation for s dot? You just need to solve for the time when the ball is at 96 feet.

 

[GunnaSix]

Where did you get the -16h term in s'(t)?

 

[morphine]

My intermediate step is: (-16t^2 - 32th - 16h^2 + 80t + 80h + 16t^2 - 80t) / h

Then I cancel out like terms, factor out the h and cancel it. Then I'm left with s'(t) = -16h -32t + 80.

Where have I gone wrong?

 

[GunnaSix]

You forgot that the derivative is the limit as h approaches 0.

 

[Phyisab****]

You left out the most important part of your limit definition!

 

[morphine]

Argh, it's these little things that kill me in math!

So the h term becomes 0, so:

s'(t) = 80 - 32t

But I'm still unclear how I get the velocity at 96 feet. I recognize that the velocity is the slope of the tangent line but... Oh wait 32 would be the slope of that line, correct? So the answer is -32??

 

[GunnaSix]

First you have to solve the position equation for t, then find the velocity at that t.

 

[Mark44]

Use the formula for s(t) to find the time t when s(t) is equal to 96. Then substitute that time into your formula for s'(t) - the velocity.

For example (and these aren't the right numbers) suppose you found that s(t) = 96 for t = 12 sec. Then you would evaluate s'(12) to get the instantaneous velocity at t = 12 sec.

 

[morphine]

Not getting anywhere with that, sorry I need another hint.

All I've been able to come out with are answers that don't make sense.

By the way, it's multiple choice, and the possible answers are: 80, -80, 32, 16, -16 feet/sec

 

[Mark44]

Show us what you're trying to do and we'll set you straight.

 

[morphine]

OK.

96 = 80t - 16t^2
96 / 16 = (80t -16t^2) / 16
6 = 5t - t^2
y = -t^2 + 5t +6
y = -1(t + 2)(t + 3)

soooo t = -2 or -3?

I feel like I am going about this the wrong way... Plugging either of those in gets a number above 100.

 

[Mark44]

OK.

96 = 80t - 16t^2
96 / 16 = (80t -16t^2) / 16
6 = 5t - t^2

So far, so good, but you're making this harder than it should be. I would have written 16t^2 - 80t + 96 = 0 as my 2nd equation here. Then I would divide by 16 on both sides to get
t^2 - 5t + 6 = 0.
This factors to (t -3)(t - 2) = 0, so t = 2 or t = 3.

y = -t^2 + 5t +6

Where did y come from?

y = -1(t + 2)(t + 3)

soooo t = -2 or -3?

I feel like I am going about this the wrong way... Plugging either of those in gets a number above 100.

From my work above, the times when the ball is at 96 ft are t = 2 and t = 3. Use your formula for velocity (s'(t)) to find the velocity at these two times. You should find that the two velocity values are numerically equal but opposite in sign.

 

[morphine]

Right you are, so I get 16 on the way up or -16 on the way down. Thanks for your help!

 

[Mark44]

Be sure to include units - ft/sec.