juggler performs under ceiling 2.4 m above his hands....

[alice]

A juggler performs in a room with a ceiling that is 2.4 m above the level of his hands. He throws a ball (diameter of 10 cm) vertically so that it just reaches the ceiling.

(a) With what initial velocity does he throw the ball?
(b) How long does it take for the ball to reach the ceiling?
(c) He throws up a second ball with the same initial velocity at the instant that the first ball is at the ceiling. How long after the second ball is thrown do the two balls pass each other?
(d) When do the balls pass each other, how far are they above the juggler’s hands?

I am uncertain where to start. i no that the ball is actually 2.3m above the level of his hand. I don't no where to go next. can someone please help

Regards
Alice

 

[mmm4444bot]

Freefalling ...

Hi Alice:

You need to explain what you've learned. There is a LOT of prerequisite knowledge required to solve this exercise.

I don't know what class you're in or what methods you've been taught.

Review the material at the following web site. Then, please post a reply to let us know if anything there looks familiar.

PurpleMath module on Freefall

Cheers,

~ Mark

 

[daon]

Re: Help with solving velocity problem

I would first find the time it takes for the ball to reach the ceiling... which would be the same amount of time for the ball to fall back into his hand. This is because the ball is under constant acceleration and must travel the same distance it did when it was on its way up,

Why do this? Because we know the initial velocity on the way down. Its zero... because the ball stops before it falls.

Use your equations:

\(\displaystyle d = d_o + v_ot + \frac{1}{2}at^2\).

The ball traveles 230cm downward (d=-230cm), use your zero of displacement at the ceiling (do=0). Initial velocity is zero (Vo=0). Acceleration (a) is -9.8m/s^2 downward. Now plug in:

\(\displaystyle -230cm = 0 + (0)t + \frac{1}{2}(-9.8m/s^2)t^2\).

Now solve for t. Once you have your time, you can find the final velocity of the ball when returing back to his hand with the following equation:

\(\displaystyle v=v_o + at\).

You'll use the time from above and the same gravity as your acceleration. This answer for velocity will have the same speed as the initial release velocity only it will be in the opposite direction (read:change the sign). The reason for this is the same as the reason for the time being the same.

 

[alice]

Re: Help with solving velocity problem

This has helped a little, however i am stuck on question c and d.

(c) He throws up a second ball with the same initial velocity at the instant that the
first ball is at the ceiling. How long after the second ball is thrown do the two balls
pass each other?

(d) When do the balls pass each other, how far are they above the juggler’s hands?

The rules i am using are
- vF = vI = at
- vF[sup:re9ms8jn]2[/sup:re9ms8jn]= vI[sup:re9ms8jn]2[/sup:re9ms8jn] = 2 ad and
- d = vIt = 1/2 at[sup:re9ms8jn]2[/sup:re9ms8jn]

if anyone can help it would be muchly appreciated.
Regards
Alice

 

[Deleted member 4993]

Re: Help with solving velocity problem

This has helped a little, however i am stuck on question c and d.

(c) He throws up a second ball with the same initial velocity at the instant that the
first ball is at the ceiling. How long after the second ball is thrown do the two balls
pass each other?

Assuming you have done the other two parts correctly

(a) With what initial velocity does he throw the ball?..............Let the answer be V[sub:23l05teo]a[/sub:23l05teo]
(b) How long does it take for the ball to reach the ceiling?..............Let the answer be t[sub:23l05teo]b[/sub:23l05teo]

Let the time when the two balls pass each other = t[sub:23l05teo]c[/sub:23l05teo]

Let the distance from hand when the two balls pass each other = d[sub:23l05teo]4[/sub:23l05teo]

As the first ball starts to fall the intial speed = 0

So

\(\displaystyle 2.3 \, - \, d_4 \, = \, \frac{9.8}{2}\cdot t_c^2\)...........................(1)

So the second ball is thorown upward with a speed V[sub:23l05teo]a[/sub:23l05teo]

\(\displaystyle d_4 \, = V_a \cdot t_c\, - \, \frac{9.8}{2}\cdot t_c^2\)...........................(2)

Add (1) and (2) - and solve for t[sub:23l05teo]c[/sub:23l05teo]

(d) When do the balls pass each other, how far are they above the juggler’s hands?

solve for d[sub:23l05teo]4[/sub:23l05teo] from (2)

The rules i am using are
- vF = vI = at
- vF[sup:23l05teo]2[/sup:23l05teo]= vI[sup:23l05teo]2[/sup:23l05teo] = 2 ad and
- d = vIt = 1/2 at[sup:23l05teo]2[/sup:23l05teo]

if anyone can help it would be muchly appreciated.
Regards
Alice