Ball thrown vertically, only the time is given

[Wanting to Learn]

Homework Statement

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A ball player catches a ball 3.33 s after throwing it vertically upward.

With what speed did he throw it?
What height did it reach?

Homework Equations

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d = v₀t + .5at²
I'm not sure whether this is one of the equations which should be used for this problem.
[I also have from class (just started AP Physics 1) a list of 5 other equations which I could try to post if that would be helpful, I'm not sure what most of them mean/how any of them could be applied (aside from d=tv).]

gravity 9.8 m/s² (pulling the ball downwards, not sure how to use this info in solving)

3.33s/2=1.665s ball spends going in each direction (up and down)

The Attempt at a Solution

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I'm not really sure how to start/what to do with this problem.
I think there need to be 2 parts, going up and going down.

d = 0 + .5*-9.8*1.665 = -8.1585
m • s = m/s
s²
I'm not sure what this number represents, if it does represent anything. I make acceleration negative because of the gravity pulling down but I'm guessing that is not how it needs to be used. Also I think this ignores the acceleration of the ball being thrown up, which was not given, so I'm very confused about what I should do.

Thank you in advance for any and all help.

 

[scottdave]

How did you arrive at your value for v0?
Do you know the equation for velocity in constant acceleration? v = v0 + a*t
You use this as well as thinking about the symmetry of the problem. If air resistance is neglected, then the trip up will be the same as the trip down.
Also in your equations, are you supposed to just multiply by time, or something else?

 

[berkeman]

d = v0t + .5at2

That's pretty close. You do want to use an equation for the vertical motion as a function of time, given some Vo upward and the acceleration of gravity downward (g = 9.8m/s^2 downward). Be sure to include the initial height (which you would set equal to the final height for the full path of the ball from throw-to-catch.

I'd prefer to use Y for the vertical height of the ball, so the equation becomes more like:
$$y(t) = y_0 + V_{0y} t - \frac{1}{2}g t^2$$
Can you keep going from there?

EDIT -- oops, scottdave beat me to it...

 

[Wanting to Learn]

ThanksOkay, to make sure I understand what the different variables all mean, is this correct?
y(t) = height (or distance gone) at time t (also known as position)
y₀ = position at time 0, is this 0 for the first part (going up?)
V₀ = velocity at time 0, this is the unknown? What is the sub y for?
t = time (seconds over which this is happening, in this case 1.665s for each part)
g = 9.8m/s²

So I can plug in

y(t)=y₀+V_0yt − 1/2 gt²

y(1.1665) = 0 + 1.1665V_0y + -.5*9.8*1.1665

=1.665V - 13.5839...
(I worked out the units on paper to be m)​

I'm not sure what to set this equal to, do I use 3.33s for time? I'm not sure that would work because it would be including the 1.1665 seconds I already calculated/accounted for?

 

[Wanting to Learn]

Thank you everyone, with your help I figured it out enough to get the correct answer.
The only thing I am still unsure of is why I have to multiply the answer I am getting for the velocity by 2 for it to be correct.
I plugged in:
x_f = x₀ +V_x0t + .5gt²
0 = 0 + 1.665V + .5*9.8*1.665²
and got V = -8.1585 which I made positive (absolute value) but then had to multiply by 2 for it to be correct , why is that?
(The correct answer is 16.3m/s according to the homework (it's online so you put in answer and it tells you if correct).)

 

[berkeman]

V0 = velocity at time 0, this is the unknown?

Yes, Vo is one of the unknowns you are asked to solve for.

What is the sub y for?

Later in your class, you will start solving problems in 2 and 3 dimensions. So you will need to keep the x, y, z components accounted for separately. This will lead into solving vector type problems. l

y(1.1665) = 0 + 1.1665V0y + -.5*9.8*1.1665
=1.665V - 13.5839...
(I worked out the units on paper to be m)I'm not sure what to set this equal to

I would start by setting y(3.33s) = y(0) = 0. That sets the "datum" or reference height at zero for the start of the throw, and for when you catch it. Then just solve for Voy. Then use the same equation with the now-known Voy to solve for the height at half-flight (half the flight time), which is the top of the ball's travel.

 

[scottdave]

Thank you everyone, with your help I figured it out enough to get the correct answer.
The only thing I am still unsure of is why I have to multiply the answer I am getting for the velocity by 2 for it to be correct.
I plugged in:
x_f = x₀ +V_x0t + .5gt²
0 = 0 + 1.665V + .5*9.8*1.665²
and got V = -8.1585 which I made positive (absolute value) but then had to multiply by 2 for it to be correct , why is that?
(The correct answer is 16.3m/s according to the homework (it's online so you put in answer and it tells you if correct).)

So the time for x_final=0 is not 1.665 seconds. At 1.665 seconds it has only gone half of the trip, so where is the ball?
If you want to use x_final = 0, then what should the time be?

 

[Wanting to Learn]

Oh okay, I think I understand now.
Thank you!