Throwing a ball: Motion in one direction

[lNVlNClBLE]

~ Throwing a ball: Motion in one direction ~

Homework Statement

The boy throws the ball vertically from the top of the building. The ball travels upward for 4 seconds. a) What was the original velocity of the ball when it was thrown? b) How high above the building does the ball go? When the ball begins its descent, it passes its original position at the top of the building. c) What is the velocity of the ball at this point? d) If the building is 40m, compute the elapsed time of flight from when the ball left the boy's hand. e) At what velocity does the ball hit the ground? f) What is the acceleration of the ball at the peak of its flight? g) What is the acceleration of the ball just before it hits the ground?

Homework Equations

v = v-original + at
Delta(x) = V-original(time) + (1/2)at^2
Ymax = V-original(time) + 1/2 at^2

The Attempt at a Solution

a) Used velocity equation instead.
0 = Vo + (-9.8)(4)
0 = Vo + (-39.2)
39.2m/s = Vo

b) This calculation makes no sense to me. EDIT: Maybe this is better. I still don't see how a boy could possibly throw a ball twice the height of a building he's standing on top of.
Ymax = Vo(time) + (1/2) at^2
Ymax = 39.2(4) + (1/2)(-9.8)(16)
Ymax = 156-78.4
Ymax = 78.4

c)
v = Vo + at
v = 39.2 + (-9.8)(8)
v = -39.2 m/s

d)
-40 = 39.2t + (1/2)(-9.8)(t)^2
-40 = 39.2t + (-4.9)(t)^2
-4.9t^2 + 39.2t + 40 = 0
4.9t^2 - 39.2t - 40 = 0

= 39.2 plus/minus sqrt (1536.64 + 784)/9.8
= 39.2 plus/minus sqrt (2320.64)/9.8
= 39.2 plus/minus 48.17/9.8
t = 8.92 seconds

WHEW.

e) v = Vo + at
v = 39.2 + (-9.8)(8.92)
v = 39.2 + (-87.42)
v = -48.22
v = 48.22 m/s

f) -9.8 m/s^2

g) -9.8 m/s^2

 

[Doc Al]

a) I assumed that displacement or delta(x) was 0, but I could be wrong.

It says the ball travels upward for 4 seconds, not that the round trip back to the start took 4 seconds.

Redo this, as it will affect the other answers.

 

[lNVlNClBLE]

It says the ball travels upward for 4 seconds, not that the round trip back to the start took 4 seconds.

Redo this, as it will affect the other answers.

I'm still not sure how this affects the question. Can I assume that since it takes 4 seconds to reach the peak, it will take 8 seconds to fall back down to its original position?

Physics noob, here.

 

[Doc Al]

I'm still not sure how this affects the question. Can I assume that since it takes 4 seconds to reach the peak, it will take 8 seconds to fall back down to its original position?

Sure. That's one way to do it. Another way is to use another kinematic equation that doesn't involve distance.

(Do it both ways!)

 

[lNVlNClBLE]

Sure. That's one way to do it. Another way is to use another kinematic equation that doesn't involve distance.

(Do it both ways!)

One more question, and thank you for the help by the way.

How am I to find the displacement, or distance, if I don't have an initial velocity? Using the equation Delta(x) = V-original(time) + (1/2) at^2, I have all of the variables besides Vo and delta(x) - and I assume delta(x) is stated implicitly somewhere in the problem and I'm just missing it -

 

[SHISHKABOB]

you don't have an initial velocity, but you do have velocity after 4 seconds. When the ball hits the peak, its velocity is equal to zero.

 

[lNVlNClBLE]

you don't have an initial velocity, but you do have velocity after 4 seconds. When the ball hits the peak, its velocity is equal to zero.

Thank you sir, it's much appreciated.

 

[Doc Al]

How am I to find the displacement, or distance, if I don't have an initial velocity? Using the equation Delta(x) = V-original(time) + (1/2) at^2, I have all of the variables besides Vo and delta(x) - and I assume delta(x) is stated implicitly somewhere in the problem and I'm just missing it -

You first solve for the initial velocity, of course. That's why it's part a!

If you use the displacement formula to solve for the initial velocity, you must choose an interval for which you already know the displacement. Such as a round trip. But you can also use the velocity formula to solve for the initial velocity.

 

[lNVlNClBLE]

You first solve for the initial velocity, of course. That's why it's part a!

If you use the displacement formula to solve for the initial velocity, you must choose an interval for which you already know the displacement. Such as a round trip. But you can also use the velocity formula to solve for the initial velocity.

Reworked, but I'm not sure if it's correct. I wish this thing came with an answer key.

EDIT2: Worked out part c.

 

[Doc Al]

a) Used velocity equation instead.
0 = Vo + (-9.8)(4)
0 = Vo + (-39.2)
39.2m/s = Vo

b) This calculation makes no sense to me. EDIT: Maybe this is better.
Ymax = Vo(time) + (1/2) at^2
Ymax = 39.2(4) + (1/2)(-9.8)(16)
Ymax = 156-78.4
Ymax = 78.4

Looks good now.

c) Will attempt after initial help - Can I just use V = V-original + at here?

Sure. (But if you catch the trick, this shouldn't require any calculation.)

d) Will attempt after initial help - Clueless as of now. I'm new to physics, is there a general formula to use for a constant increasing downwards acceleration? 1/2gt^2?

Use the same formula. What's the displacement from the initial position?

 

[lNVlNClBLE]

Looks good now.Sure. (But if you catch the trick, this shouldn't require any calculation.)

Use the same formula. What's the displacement from the initial position?

Woot.

I'm assuming the trick is some sort of law that states if there's no other resistance, an object that travels x meters upward at a velocity will have the same velocity as the initial if it falls the same x meters?

39.2? Which formula would I be plugging this into?

 

[Doc Al]

I'm assuming the trick is some sort of law that states if there's no other resistance, an object that travels x meters upward at a velocity will have the same velocity as the initial if it falls the same x meters?

Yep. The same speed, not velocity, since the direction is opposite. (Thus the minus sign.)

 

[Doc Al]

I still don't see how a boy could possibly throw a ball twice the height of a building he's standing on top of.

That depends on how fast the boy can throw and how high the building is, doesn't it? Of course, this kid is obviously a major league fastball pitcher! (Not very realistic.)

f) Assuming this is just 0m/s^2, since it should have no acceleration at the absolute peak.

 

[lNVlNClBLE]

That depends on how fast the boy can throw and how high the building is, doesn't it? Of course, this kid is obviously a major league fastball pitcher! (Not very realistic.)

Lol. *facepalm*

Was I wrong? I'm still trying to figure out which equation I'd use to solve part d. Any hints? I suppose the displacement at the peak is 39.2, but I don't know how that helps me.

 

[SHISHKABOB]

an important thing about the equations used here is that they only work when acceleration remains constant. Also, acceleration is caused by forces, which means basically that acceleration only happens when something pushes on something else. In this situation, after the ball leaves the kid's hand, there is nothing pushing on the ball except for gravity (if we ignore air resistance). This is also why they make the important distinction of "just before" in g.

 

[lNVlNClBLE]

an important thing about the equations used here is that they only work when acceleration remains constant. Also, acceleration is caused by forces, which means basically that acceleration only happens when something pushes on something else. In this situation, after the ball leaves the kid's hand, there is nothing pushing on the ball except for gravity (if we ignore air resistance). This is also why they make the important distinction of "just before" in g.

Another thing I was trying to understand; I know that when objects fall in a non-vacuum, their acceleration is not constant. AKA, after one second, a ball would NOT fall 9.8 meters, but something like 4.

Arg.

 

[Doc Al]

Was I wrong?

About the acceleration being zero? Yep, way wrong. What's the acceleration of a projectile at any point in its motion?

I'm still trying to figure out which equation I'd use to solve part d. Any hints? I suppose the displacement at the peak is 39.2, but I don't know how that helps me.

You'd use the same equation as before. (The second in your list.) Displacement (Δx) means: Final position - initial position.

 

[SHISHKABOB]

Another thing I was trying to understand; I know that when objects fall in a non-vacuum, their acceleration is not constant. AKA, after one second, a ball would NOT fall 9.8 meters, but something like 4.

Arg.

that's because air is in the way, basically. It's the same idea as why it's hard to move through water. The water is "pushing" back at you in the form of resistance. Air "pushes" on falling objects in the same way.

in simple problems like yours, we just forget about it because it makes things a lot more complicated

 

[Doc Al]

Another thing I was trying to understand; I know that when objects fall in a non-vacuum, their acceleration is not constant. AKA, after one second, a ball would NOT fall 9.8 meters, but something like 4.

First solve the 'simple' problems that ignore complications such as air resistance--you can worry about the harder and more realistic problems later.

 

[lNVlNClBLE]

About the acceleration being zero? Yep, way wrong. What's the acceleration of a projectile at any point in its motion?

You'd use the same equation as before. (The second in your list.) Displacement (Δx) means: Final position - initial position.

I don't know. ._. I know the acceleration in the x direction is always constant, I don't know about the y direction.

Will try.

 

[Doc Al]

I don't know. ._. I know the acceleration in the x direction is always constant, I don't know about the y direction.

The only direction of motion here is along the vertical axis. (Calling it x or y doesn't matter here.)

 

[lNVlNClBLE]

The only direction of motion here is along the vertical axis. (Calling it x or y doesn't matter here.)

That much I understand. I was always under the impression that any object at its peak would have both an acceleration and velocity of zero, since said object has stopped moving before it switches direction.

Did some work for part d using the displacement formula.
Does my algebra fail, or is that unsolvable?

 

[Doc Al]

That much I understand. I was always under the impression that any object at its peak would have both an acceleration and velocity of zero, since said object has stopped moving before it switches direction.

Nope. The velocity is momentarily zero, but the acceleration remains constant throughout the motion.

Did some work for part d using the displacement formula.
Does my algebra fail, or is that unsolvable?

It's a quadratic equation. Perfectly solvable.

d)
78.2 = 39.2t + (1/2)(-9.8)t^2
78.2 = 39.2t + (-4.9)t^2

Right equation, but wrong displacement.

 

[lNVlNClBLE]

Nope. The velocity is momentarily zero, but the acceleration remains constant throughout the motion.

It's a quadratic equation. Perfectly solvable.

Right equation, but wrong displacement.

Does the acceleration require calculation, then, or is there something I'm required to know?

Oh, I'll try to get that in standard quadratic form and then solve it.

Which displacement should I be using? Arg. If it reaches 78.4m above the building, isn't the displacement 78.4?

 

[Doc Al]

Does the acceleration require calculation, then, or is there something I'm required to know?

You just have to know. What's the acceleration due to gravity?

Oh, I'll try to get that in standard quadratic form and then solve it.

Yep.

Which displacement should I be using? Arg. If it reaches 78.4m above the building, isn't the displacement 78.4?

That would be the displacement from the start to the top of the motion. What you want is the displacement for the entire trip, from start to when it hits the ground.

 

[lNVlNClBLE]

You just have to know. What's the acceleration due to gravity?

Yep.

That would be the displacement from the start to the top of the motion. What you want is the displacement for the entire trip, from start to when it hits the ground.

9.8.

Once I figure out my displacement, I'll get back to you. Lol.

><. I'm so stumped. How do I find the displacement for the whole trip; I'm still trying to figure out what I need to plug in for my displacement rather than 78.4 or 39.2

 

[Doc Al]

How do I find the displacement for the whole trip; I'm still trying to figure out what I need to plug in for my displacement rather than 78.4 or 39.2

What's the initial position? What's the final position? (Measure from the same point, of course.)

 

[lNVlNClBLE]

What's the initial position? What's the final position? (Measure from the same point, of course.)

Starts at 40m, goes up to 118.4, then drops 118.4 to the ground.

0-40 = -40?

 

[Doc Al]

Starts at 40m, goes up to 118.4, then drops 118.4 to the ground.

0-40 = -40?

Good!

I'd say: Starts at 40 m above the ground and ends up at 0 m above the ground. (No need to worry about the intermediate steps.)

 

[lNVlNClBLE]

Good!

I'd say: Starts at 40 m above the ground and ends up at 0 m above the ground. (No need to worry about the intermediate steps.)

Phew. Long time since I used the quadratic. Can you check to see if that's right?
Also, I've never used it before in physics, so I'm assuming the whole negative part of the plus/minus can be ignored since there's no such thing as negative time.

EDIT: Completed the rest of the problem with that time. I assume f is also just 9.8 m/s^2. If so, are those accelerations -9.8, or just 9.8?