Grade 11 Physics Question: Vertical Throwing and Falling Balls

[w3tw1lly]

Heres question I need help with. I'm in Grade 11 Physics.

A ball is thrown vertically upward from a window that is 3.6 m above the ground. The ball's initial speed is 2.8 m/s and the acceleration due to gravity is 9.8 m/s^2.

a) What is the ball's speed when it hits the ground?
b) How long after the first ball is thrown should a second ball be simply dropped form the same window so that both balls hit the ground at the same time?


I don't know where to start. I have tried an equation but it keeps seeming to be the wrong answer. I get approx 8.9 m/s [down] for a).

 

[G01]

What equation have you tried? It's hard (and against the rules) to help when you haven't shown any work, since I don't know where you are getting stuck in the problem.

 

[w3tw1lly]

V2^2 = V1^2 + 2*a*dV2 = Final speed
V1 = Initial speed

The thing I'm having a problem with is if you say that + = up... That makes it so you have to find a square root of a negative number which you can't do. The only way it works is if 3.6 becomes negative too but I don't think distances can go negative?

 

[G01]

The change in distance can be negative and is here. Remember that, since you have defined up as the positive direction, the ball is moving 3.6 m in the negative direction. This means that it would have changed it's position by -3.6m. In other words, the ball ends up 3.6m below where it started.

Does this make sense?

Also, remember what direction V2 is pointing in once you find it.

 

[cristo]

V2^2 = V1^2 + 2*a*d


V2 = Final speed
V1 = Initial speed

The thing I'm having a problem with is if you say that + = up... That makes it so you have to find a square root of a negative number which you can't do. The only way it works is if 3.6 becomes negative too but I don't think distances can go negative?

If you take upwards as being positive then the displacement will be -3.6; i.e. the overall displacement of the ball will be 3.6 metres downwards.

 

[w3tw1lly]

So then 8.9 m/s is right...Thanks I just wasn't really sure about that, I kinda joined the course late so I'm not too bright right now. Thanks again.

 

[cristo]

So then 8.9 m/s is right?

Show your working! It's a lot easier to check your work if you show the equation you have used as the first step, and then substitute in the numbers. You wouldn't just say "8.9" in an exam now, would you?

 

[G01]

So then 8.9 m/s is right...Thanks I just wasn't really sure about that, I kinda joined the course late so I'm not too bright right now. Thanks again.

Cristo is right. We would have been able to tell you if the answer was correct much more quickly if you showed your work from the beginning!

 

[w3tw1lly]

In this case + = down

V2^2 = (2.8)^2 + 2(9.8)(3.6)
= 7.84 + 70.56
= 78.4
V2 = sqrt(78.4)
V2 = 8.8543...

Therefore the speed of the ball as it hits the ground is 8.9 m/s.Sorry for bad form.

 

[G01]

In this case + = down

V2^2 = (2.8)^2 + 2(9.8)(3.6)
= 7.84 + 70.56
= 78.4
V2 = sqrt(78.4)
V2 = 8.8543...

Therefore the speed of the ball as it hits the ground is 8.9 m/s.Sorry for bad form.

That is correct. Just remember that down is usually taken to be negative, so you may need to change the sign if you have to hand this in. Either that, or explicitly state that down is positive like you did here.

EDIT: I wasn't thinking and I almost gave away the answer there in my above post. I edited the post, and it seems found it yourself anyway. Good Job!

 

[cristo]

In this case + = down

V2^2 = (2.8)^2 + 2(9.8)(3.6)

With your sign conventions, this should be negative. It doesn't affect the answer though.

= 7.84 + 70.56
= 78.4
V2 = sqrt(78.4)
V2 = 8.8543...

Therefore the speed of the ball as it hits the ground is 8.9 m/s.

Correct

Sorry for bad form.

Don't worry about it. Just show your work next time and it'll be easier to help!

Ok, I'll stop butting in now. Sorry G01!

 

[w3tw1lly]

Ok I think I'm stuck on the second part now.

I have calculated that it will take 0.857142857 seconds for a ball to travel if it were to be just dropped from 3.6 m above ground. I don't even know what equation I should use now to find out at what time I should drop the ball at. I think I have figured that the ball that is thrown up takes 0.617793617 seconds to fall after it reaches the point it was thrown. I just can't find how long it was up for? Nevermind I think it is 0.75 seconds. Sorry I'm all over the place.

 

[G01]

Ok I think I'm stuck on the second part now.

I have calculated that it will take 0.857142857 seconds for a ball to travel if it were to be just dropped from 3.6 m above ground. I don't even know what equation I should use now to find out at what time I should drop the ball at. I think I have figured that the ball that is thrown up takes 0.617793617 seconds to fall after it reaches the point it was thrown. I just can't find how long it was up for? Nevermind I think it is 0.75 seconds. Sorry I'm all over the place.

To find how long the ball travels for when it is thrown up, try this. Find how long it takes the ball to reach its highest point. Then find how long it takes for the ball to fall to the ground from the highest point. Then add them up.

Then you will know how long it takes both balls to hit the ground. You should then be able to reason how long after you should wait before dropping the second ball.

Ok, I'll stop butting in now. Sorry G01!

Don't worry Cristo, I'm a forgiving kind of guy.

 

[w3tw1lly]

Ok so I'm going to do:

t = (V2 - V1) / a
= (2.4 - 0) / 9.8
= 2.4 / 9.8
= 0.244897959

Since that is only how long it takes for it to go up to its max height, it must be multiplied by two.
0.489795918

Ok then I add that onto how long it takes to fall from 3.6 metres.

0.489... + 0.617... = 1.107...

Ok so that is how long it takes for the thrown one to fall. The dropped one takes 0.857...

So to figure out what time to drop the dropped one at so that it hits the ground at the same time the thrown one does you have to subtract it.

1.107... - 0.857... = 0.25


Therefore the second ball should be dropped at .25 seconds so that it hits the ground at the same as the thrown ball.

 

[G01]

Your reasoning is completely correct but you seem to have made a mistake in the initial velocity.

The initial velocity of the first ball is 2.8m/s not 2.4m/s. This should change your answer a little bit.

Other than that, everything seems good.

 

[w3tw1lly]

Thank you for noticing that discrepancy and thank you both for the help it was very useful. Hopefully the teacher will help me catch up.