Does this seem correct? (throwing a tennis ball straight up and then catching it)

  • #1
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Summary:
Does this seem correct
Hello I’m hoping someone could help

I was asked by a fellow student the following question

A 6 foot tall boy is stood in his schools gymnasium, he throws a standard full-size tennis ball 🎾 up in the air to the height of 20 feet and it lands back into his hand at the same distance from the floor that he threw it from (air resistance negligible)

How long in seconds would it take from leaving his hand to being caught again?


I have come to the answer of approx 5.5 seconds


Can someone please confirm this to be correct or close? (I can provide my working out if needed or requested)

Thanks 😊
 

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  • #2
berkeman
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Welcome to PF. :smile:

Can you show the equations you used to figure this out? I got a different answer, but I just did a quick calculation on a Post-It note, and ran out of room near the end...
 
  • #3
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Thanks for the fast response 😃

Sorry if this is messy lol


Here’s my workout out

As acceleration of the throw is not stated, I will assume it is 0.5m/s is shall name "a".

20 ft is 6 meters which shall be labeled "S"

I want to find time in seconds. I shall label "t"

Equation required:
S = ut + ½at²

To throw a ball at 0.5m/s at a distance of 6 meters, this means it will take 12 seconds. Let's call this t.

Now using the formula.
Where u is 0 indicated to be initial state of rest.
t is 12 s
a is 0.5 m/s
S = ut + at
6 = 0.5 X 12

6 meters satisfied provided an initial speed of 0.5m/s at the throw.

So using this...

V = u + at² is to be used

V = final velocity

Which is 0.5m/s.

So rearranged for t the equation gives:

(v - u) x a / 2

= (0.5 - 0) x 9.81 / 2
= 2.45seconds

This is going up

So 2.45 seconds coming down

5.5 seconds

Which roughly agrees with the previous formula
Remember going up and down involves using 9.81 for gravity

Takes 5.5 seconds
 
  • #4
hutchphd
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Summary:: Does this seem correct

How long in seconds would it take from leaving his hand to being caught again?

You need to know the height of the ball release(and catch) to get this answer. Suppose it is 4ft (the numbers work out nicely). You need assume nothing else. Give it a try.
 
  • #5
DaveC426913
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As acceleration of the throw is not stated, I will assume it is 0.5m/s is shall name "a".
The acceleration of the throw is not stated because it does not factor in. The only acceleration you need to worry about is gravity.
 
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  • #6
berkeman
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You need to know the height of the ball release(and catch) to get this answer. Suppose it is 4ft (the numbers work out nicely). You need assume nothing else. Give it a try.
He did state the max height and the difference between the starting and ending heights (0). My problem was that I misread it as 20m, not 20 feet... :smile:
Summary:: Does this seem correct

A 6 foot tall boy is stood in his schools gymnasium, he throws a standard full-size tennis ball 🎾 up in the air to the height of 20 feet and it lands back into his hand at the same distance from the floor that he threw it from
 
  • #7
hutchphd
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Summary:: Does this seem correct

he throws a standard full-size tennis ball 🎾 up in the air to the height of 20 feet

He did state the difference between the starting and ending height.

Symantics again. Isn't the height of a basketball hoop 10 ft? In my vernacular height is distance from the flat earth. :world:🐘🐘
 
  • #8
berkeman
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If this 6 foot boy is tossing a tennis ball up to a basketball hoop, he's got bigger problems... :wink:
 
  • #9
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Just tried again, 2.2 can’t be right....where am I going wrong hahaha

s = ut + ½at

s = 6m
u = 0
a = 9.81 m/s²

Rearrange gives:

s / ½g = t²

6/4.4 = 1.3²

Root = 1.1 s up
1.1 seconds down.

2.2 seconds
 
  • #10
hutchphd
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16.0 ft height change would be 2.00 seconds. So I can buy the answer.....
 
  • #11
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You need to know the height of the ball release(and catch) to get this answer. Suppose it is 4ft (the numbers work out nicely). You need assume nothing else. Give it a try.
My bad 🤦🏻

The question was proposed meaning the ball travels 20 feet from release, so can disregard the distance of the boys hand to the ground :)

Anyone able to offer an answer and short explanation so I can see where I went wrong

Thanks 🙏
 
  • #13
hutchphd
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t+t =/= t2 (esp. if u=0)
##t^2## is correct (typo I think)
2.2 seconds
This seems correct to me ( a little more than what you get for 16 ft, right?)

.
 
  • #14
PeroK
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Just tried again, 2.2 can’t be right....where am I going wrong hahaha

s = ut + ½at

s = 6m
u = 0
a = 9.81 m/s²

Rearrange gives:

s / ½g = t²

6/4.4 = 1.3²

Root = 1.1 s up
1.1 seconds down.

2.2 seconds
You can do it this way. We take upwards to be positive, so ##a = -g##. First, we look at the total displacement: $$s = ut - \frac 1 2 gt^2$$
When the ball returns to its starting point we have ##s = 0##, so:
$$u = \frac 1 2 gT$$ where ##T## is the total time in the air. And, rearranging this we have:
$$T = \frac{2u}{g}$$ Now we need to find a relationship between maximum height above the launch point, ##h##, and initial velocity ##u##. Note that the maximum height occurs when the speed is zero. We use the equation for velocity of the ball:
$$v = u - gt$$And, setting ##v = 0## we have ##t = \frac u g##, where ##t## is the time to reach the maximum height.

Note that ##T = 2t##. And we see that the time to go up and back down to the starting point is always twice the time to reach the maximum height.

Now we can calculate the maximum height in terms of ##T##:
$$h = u(\frac T 2) - \frac 1 2 g(\frac T 2)^2 = \frac 1 2 gT(\frac T 2) - \frac 1 2 g(\frac T 2)^2 = \frac{gT^2}{8}$$Where we used ##u = \frac 1 2 gT##.

Finally, we rearrange that equation to get: $$T^2 = \frac{8h}{g}$$

I'm used to SI units, but if we use American units, with ##g \approx 32 ft/s^2##, then we have:
$$T = \frac{\sqrt h}{2} = \sqrt{\frac h 4}$$ Where ##T## is measured in seconds and ##h## is measured in feet. In your example, we have ##h = 20ft##, hence: $$T = \sqrt 5 \ s = 2.2 s$$
 
  • #15
PeroK
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PS you can also see that if ##T = 5.5s## then ##h = 120ft##. It's interesting that you had little intuition from experience (watching sports, for example) of just how high you would need to throw a ball for it to stay in the air for over ##5## seconds!
 
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  • #16
rude man
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Summary:: Does this seem correct

Hello I’m hoping someone could help

I was asked by a fellow student the following question

A 6 foot tall boy is stood in his schools gymnasium, he throws a standard full-size tennis ball 🎾 up in the air to the height of 20 feet and it lands back into his hand at the same distance from the floor that he threw it from (air resistance negligible)

How long in seconds would it take from leaving his hand to being caught again?


I have come to the answer of approx 5.5 seconds

Can someone please confirm this to be correct or close? (I can provide my working out if needed or requested)

Thanks 😊
The problem can't be solved unless you specify the height at which the ball was thrown (where his hand was when he let it go)..

Hint: it takes as long for the ball to reach 20ft. as it does for the ball to return from that height back to his hand.
 
  • #17
DaveC426913
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The problem can't be solved unless you specify the height at which the ball was thrown (where his hand was when he let it go)..
Good point!

But you're also right about the "you specify". The student could simply assume a value (say, six feet) and then explicitly state that assumption as a proviso to their answer.
 
  • #18
rude man
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Good point!

But you're also right about the "you specify". The student could simply assume a value (say, six feet) and then explicitly state that assumption as a proviso to their answer.
You are exactly right!
 
  • #19
sophiecentaur
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PS you can also see that if ##T = 5.5s## then ##h = 120ft##. It's interesting that you had little intuition from experience (watching sports, for example) of just how high you would need to throw a ball for it to stay in the air for over ##5## seconds!
No actual sports experience is necessary. Just throw something in the air and start counting. We used to play a game, to see how many times you could clap; your hands between throwing a ball up and catching it. You can't clap very many times; try it.

I'm a great advocate of experiments whenever possible - especially for 'ball=park' figures.

This thread is worrying me a bit. We seem to be actually discussing whether the SUVAT equations are in question.
 
  • #20
hutchphd
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I was only worried about using intuition as a check and what the definition of "height" is. Does seem like a lot of flopping about.
 
  • #21
sophiecentaur
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I was only worried about using intuition as a check and what the definition of "height" is. Does seem like a lot of flopping about.
The original set question is clearly not aimed at any extra assumptions about experimental details. (That's what elementary text books do to the student.) Too much thinking about the problem (à la PF) is not always healthy. That's why I suggested going outside and throwing something up in the air. The answer would lie very much in the region of the (any) result from that experiment.
PF has an unfortunate habit of presenting wall to wall maths and technical asides to students who may already be totally unsure of themselves. We don't ever want to give them the answer straight, of course. A way into the problem is what's needed.
 
  • #22
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As acceleration of the throw is not stated, I will assume it is 0.5m/s is shall name "a".
"a" is a misleading name for a variable -- 0.5 m/s is a velocity, not an acceleration. As soon as the ball leaves the thrower's hand, its only acceleration will be negative (a deceleration) due to gravity.
To throw a ball at 0.5m/s at a distance of 6 meters, this means it will take 12 seconds. Let's call this t.
This makes no sense -- the ball won't be travelling at a constant velocity.
Isn't the height of a basketball hoop 10 ft?
There was no mention of a basketball hoop (or rim, as I believe they're usually called) in the problem.
 
  • #23
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The acceleration of the throw is not stated because it does not factor in.
Further, not only does it not factor in, "acceleration of the throw" isn't really a thing. The only factors are the initial height of the ball and its initial velocity when thrown. Once it leaves the thrower's hand, its only acceleration is due to gravity.
 
  • #24
sophiecentaur
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a is 0.5 m/s
This was what you called "the acceleration of the throw". That's a bit like the "force of the throw" that's a common mis-used phrase. We're only dealing with a simple free-fall situation with only g involved.
@PeroK gave you the full workings with the 'right' answer. Did you see how it all works out?
 
  • #25
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[Mentor Note -- this solution post is being allowed because the OP seems to have lost interest in the thread]

These things are best thought about by twisting the problem around to make it the easiest. You've mentioned a kid, and mentioned 20 feet, throwing and catching. That leaves some room for interpretation. Take a ball, dropped from a velocity of zero in normal gravity (or falling back down after being thrown up!). We are dealing in imperial, so it falls at 32 feet per second squared. The distance it travels is 0.5 x (32) x ( t squared). Now, you want t. So how far did it fall? Did the kid release it a head height (6 feet) and it went 20 feet up from the floor (net 14 feet)? or a foot below his head (5 feet) netting 15 feet? Did it travel up 20 feet from the release point (net 20 feet)? Once you decide how far it traveled, multiply that by 2, divide by 32, and then take the square root.

That's the time coming down.
And it took that long going up.
So multiply by two.
And that's your answer.

14 feet net --> 1.87 s
15 feet net --> 2.07 s
20 feet net --> 2.24 s
 
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