Solving for Time When a Ball is Dropped 85 Feet

[hatelove]

Homework Statement

A ball is dropped that's 85 feet above the ground. The ball's initial speed is 5 feet/second, and the acceleration of the ball is 32 feet/sec^2. How long does the ball travel before hitting the ground?

Homework Equations

distance = rate * time

The Attempt at a Solution

So the speed of the ball I have as:

[t = time]

-5 - 32t

Which I believe makes sense because the ball is initially traveling @ -5 ft/sec, and after 1 second, the ball is traveling -37 ft/sec, then after 2 seconds the ball is traveling @ -69 ft.sec, etc.

I assume we use d = rt (which is distance = rate * time).

So the total distance the ball travels is -85 feet which is equal to the total rate of -5 - 32t * time. The equation I set up like this:

-85 = (-5 - 32t) * t

I set up the quadratic like so:

-32t^2 - 5t + 85 = 0

I solved it and got:

(1/64) * (sqrt(10905) - 5) which is approx. 1.55 seconds.

But the solution says it's 2.15 seconds, and I found out that it's because the solution needs to have taken the average velocity or something, meaning the average of:

-5 ft/sec and (-5 ft/sec - (32 ft/sec(t)))

So the average turns out to be (-5 ft/sec - 16 ft/sec * t) and I can go on with the rest of the solution and I get about 2.15 seconds as the answer.

What I don't understand is why we take the average velocity of the initial and the relative velocities.

I thought -5 - 32t was relative to the distance and time that the ball had traveled. For example:

At 0 seconds:
d = (-5 - 32(0))(0)
d = (-5 - 0))(0)
d = (-5 ft/sec))(0 seconds)
d = 0 ft.

So while the ball has not yet traveled, it has a velocity of -5 ft/sec.

After 1 second:
d = (-5 - 32(1))(1)
d = (-5 - 32)(1)
d = (-37 ft/sec)(1 second)
d = -37 ft.
After 1 second, the ball was traveling at -37 ft/sec, and has traveled -37 feet towards its destination so far.

After 2 seconds:
d = (-5 - 32(2))(2)
d = (-5 - 64)(2)
d = (-64 ft/sec)(2 seconds)
d = -128 ft.
The ball is traveling at -64 ft/sec but has already past the -85 ft. mark and presumably through the ground, -43 feet more underground (but that doesn't count right now in this scenario).

So when the ball has reached its final destination after 1.55 seconds:
d = (-5 - 49.6(1.55))(1.55)
d = (-81.88 ft/sec)(1.55 seconds)
d = -84.63 ft.
The ball was traveling -81.88 ft/sec right when it hit the ground -85 feet down.

Just for example's sake, I'll calculate the math for when the ball was halfway to its mark:

(-85ft / 2) = (-5 - 32(t))(t)
-42.5 ft. = -32t^2 - 5t
0 = -32t^2 - 5t + 42.5
t = 1.07696 seconds

So after 1.07696 seconds, the ball was halfway to the ground, and:

d = (-5 - 32(1.07696))(1.07696)
d = (-39.46272 ft/sec)(1.07696 seconds)
d = -42.4997709 feet

Which means that halfway to the target (-42.5 feet down), the ball had a velocity of -39.46272 ft/sec at 1.07696 seconds after the ball started to travel.

So as a recap:

At 0 feet = ball is traveling at -5 ft/sec after 0 seconds of traveling
At -37 feet = ball is traveling at -37 ft/sec after 1 second of traveling
At -42.5 feet = ball is traveling at -39.5 ft/sec after 1.07696 seconds of traveling
At -85 feet = ball is traveling at -81.88 ft/sec after 1.55 seconds of traveling

So what is the purpose of taking the average velocity?

 

[PeterO]

Homework Statement

A pitcher throws a ball towards a wall that's 85 feet away from him. The ball's initial speed is 5 feet/second, and the acceleration of the ball is 32 feet/sec^2. How long does the ball travel before hitting the wall?

Homework Equations

distance = rate * time

The Attempt at a Solution

So the speed of the ball I have as:

[t = time]

5 + 32t

Which I believe makes sense because the ball is initially traveling @ 5 ft/sec, and after 1 second, the ball is traveling 37 ft/sec, then after 2 seconds the ball is traveling @ 69 ft.sec, etc.

I assume we use d = rt (which is distance = rate * time).

So the total distance the ball travels is 85 feet which is equal to the total rate of 5 + 32t * time. The equation I set up like this:

85 = (5 + 32t) * t

I set up the quadratic like so:

32t^2 + 5t - 85 = 0

I solved it and got:

(1/64) * (sqrt(10905) - 5) which is approx. 1.55 seconds.

But the solution says it's 2.15 seconds, and I found out that it's because the solution needs to have taken the average velocity or something, meaning the average of:

5 ft/sec and (5 ft/sec + (32 ft/sec(t)))

So the average turns out to be (5 ft/sec + 16 ft/sec * t) and I can go on with the rest of the solution and I get about 2.15 seconds as the answer.

What I don't understand is why we take the average velocity of the initial and the relative velocities.

I thought 5+32t was relative to the distance and time that the ball had traveled. For example:

At 0 seconds:
d = (5 + 32(0))(0)
d = (5 + 0))(0)
d = (5 ft/sec))(0 seconds)
d = 0 ft.

So while the ball has not yet traveled, it has a velocity of 5 ft/sec.

After 1 second:
d = (5 + 32(1))(1)
d = (5 + 32)(1)
d = (37 ft/sec)(1 second)
d = 37 ft.
After 1 second, the ball was traveling at 37 ft/sec, and has traveled 37 feet towards its destination so far.

After 2 seconds:
d = (5 + 32(2))(2)
d = (5 + 64)(2)
d = (64 ft/sec)(2 seconds)
d = 128 ft.
The ball is traveling at 64 ft/sec but has already past the 85 ft. mark and presumably through the target, 43 feet more than its intended distance (but that doesn't count right now in this scenario).

So when the ball has reached its final destination after 1.55 seconds:
d = (5 + 49.6(1.55))(1.55)
d = (81.88 ft/sec)(1.55 seconds)
d = 84.63 ft.
The ball was traveling 81.88 ft/sec right when it hit the target 85 feet away.

Just for example's sake, I'll calculate the math for when the ball was halfway to its mark:

(85ft / 2) = (5 + 32(t))(t)
42.5 ft. = 32t^2 + 5t
0 = 32t^2 + 5t - 42.5
t = 1.07696 seconds

So after 1.07696 seconds, the ball was halfway to its target, and:

d = (5 + 32(1.07696))(1.07696)
d = (39.46272 ft/sec)(1.07696 seconds)
d = 42.4997709 feet

Which means that halfway to the target (42.5 feet in), the ball had a velocity of 39.46272 ft/sec at 1.07696 seconds after the ball started to travel.

So as a recap:

At 0 feet = ball is traveling at 5 ft/sec after 0 seconds of traveling
At 37 feet = ball is traveling at 37 ft/sec after 1 second of traveling
At 42.5 feet = ball is traveling at 39.5 ft/sec after 1.07696 seconds of traveling
At 85 feet = ball is traveling at 81.88 ft/sec after 1.55 seconds of traveling

So what is the purpose of taking the average velocity?

Whoa!

see red above.

You don't have to be a major league pitcher to throw a ball at 5 ft/s [ just a couple of miles per hour - major league pitchers probably throw at closer to 100 mph] so even you can do it - even underarm like a soft-ball pitch].

OK so get a ball and throw in at 5 ft/s, then watch it accelerate to 37 ft/s in the first second after it leaves your hand!

Now don't go cheating by going to the top of a tall building and throw the ball straight down at 5 ft/s or you might actually succeed!

I did not read past that sentence I high-lighted red, as I assume you have either gone further astray, or based all you calculations on that first incorrect sentence.

 

[hatelove]

Okay, well even if it made more physical "sense" by dropping it, the numbers would still be the same except the negative of the numbers. It wouldn't change the calculations, I've tested those too. But if preferred I'll just change the wording of the problem if that's an issue...

 

[PeterO]

Okay, well even if it made more physical "sense" by dropping it, the numbers would still be the same except the negative of the numbers. It wouldn't change the calculations, I've tested those too. But if preferred I'll just change the wording of the problem if that's an issue...

It certainly makes more sense if you are dropping/trowing it down. The first problem you mentioned was a standard projectile motion question - with entirely different answers.

What makes you think that after the ball has accelerated for 1 second - to reach a speed of 37 ft/s, it won't continue to accelerate, reaching perhaps 69 ft/s after 2 seconds [unless it has already run into the ground.

I think the formula you are after is

d = v_ot + 0.5at²

Not sure how all those other calculations you were doing came about.

After all you were after the time, and was given the distance, the acceleration and the initial velocity, so from the 5 equations describing motion with constant acceleration you were after the only one that didn't involve the final velocity - because you don't know the final velocity.

distance covered after whole seconds:

1 second 21 ft
2 seconds 74 ft
3 seconds 159 ft

so a time a bit over 2 seconds sounds pretty good.

 

[PeterO]

Okay, well even if it made more physical "sense" by dropping it, the numbers would still be the same except the negative of the numbers. It wouldn't change the calculations, I've tested those too. But if preferred I'll just change the wording of the problem if that's an issue...

When trowing the ball down from a building, it is common to define "down" as "positive".
That way initial velocity, acceleration and displacement are all positive - because they are all down.

 

[hatelove]

It certainly makes more sense if you are dropping/trowing it down. The first problem you mentioned was a standard projectile motion question - with entirely different answers.

What makes you think that after the ball has accelerated for 1 second - to reach a speed of 37 ft/s, it won't continue to accelerate, reaching perhaps 69 ft/s after 2 seconds [unless it has already run into the ground.

I thought my calculations showed that after 1 second when the ball has reached a speed of 37 ft/s, at 1.07696 seconds the ball has accelerated to a speed of 39.5 ft/s, and at 1.55 seconds the speed has accelerated to 81.88 ft/s.

I think the formula you are after is

d = v{sub]o[/sub]t + 0.5at²

Not sure how all those other calculations you were doing came about.

What is this formula? I'm not sure how you got it.

After all you were after the time, and was given the distance, the acceleration and the initial velocity, so from the 5 equations describing motion with constant acceleration you were after the only one that didn't involve the final velocity - because you don't know the final velocity.

distance covered after whole seconds:

1 second 21 ft
2 seconds 74 ft
3 seconds 159 ft

so a time a bit over 2 seconds sounds pretty good.

I thought the equation of the velocity (5 ft/s + 32 ft/s * t) was universally applicable to any distance and time because the velocity increases as time passes and distance grows?

 

[PeterO]

I thought my calculations showed that after 1 second when the ball has reached a speed of 37 ft/s, at 1.07696 seconds the ball has accelerated to a speed of 39.5 ft/s, and at 1.55 seconds the speed has accelerated to 81.88 ft/s.


What is this formula? I'm not sure how you got it.


I thought the equation of the velocity (5 ft/s + 32 ft/s * t) was universally applicable to any distance and time because the velocity increases as time passes and distance grows?

That last formula you are quoting is

v_f = v_o + at

Which tells you how fast things are going, not how far they have gone.

more reply soon - have to do another job.

 

[fishspawned]

now I'm not sure what grade you are but here it is in a nutshell:

your equation: Vo + at is a measurement of the speed at any time

so you would describe that as dD/dt [where i am labeling D as the distance traveled]

assuming this you can integrate by:

dD/dt = Vo + at

∫dD = ∫Vo + at dt

therefore

D = Vot + 1/2at^2

and you can use this to determine time - but it will require solving a quadratic equation

 

[PeterO]

What is this formula? I'm not sure how you got it.

That formula is one of the set:

v_f = v_o + at

d = v_ot + 0.5at²

d = v_ft - 0.5at²

d = (v_o + v_f).t/2

v_f² = v_o² + 2ad

where

v_o = Initial velocity

v_f = final velocity

a = acceleration

d = displacement

t = time interval

As you can see, each of the variables in turn is missing from an equation.
These equations are for motion at constant acceleration.

They are all derived from a simple velocity vs time graph - which is why many problems can be solved more simply using a graph.