Advance Mechanics Question

[DanielSmith]

A pinata was projected vertically upwards from a height of 1 metre and remains in the air for 6.1 seconds.

(a) What speed was the pinata thrown at, and
(b) What is the distance the pinata travelled before returning to height of 1 meter?

[Robot10]

Was the answer for a) 29.73m/s and b)90.2m

[fun_jirachi]

Hey there!

I'm going to go through an answer with one really big assumption (because your question wasn't 100% clear, and I want to make a few clarifications) and also please tell us which part you don't understand so we can help you out!

Okay, so the main assumption here is that the base height is one metre ie. it finishes its projectile motion at an altitude of 1m. This simplifies things a lot

Subbing into the equation v=u+at, knowing that v is zero at the vertex and acceleration is -9.8 m/s/s, and also that time of flight to the vertex is half of 6.1, you can rearrange to get that:

And gets you an answer of 29.89m/s, or just 30m/s to one significant figure.

Using this information, sub it into s=ut+(1/2)at^2, using the fact that u=29.89, t=3.05, a=-9.8. The distance going to the vertex is 45.58225m, but since it comes back down we double it to get 91.1645m. To two significant figures its just 91m.

Hope this right and hope this helps

[DanielSmith]

Hey there!

I'm going to go through an answer with one really big assumption (because your question wasn't 100% clear, and I want to make a few clarifications) and also please tell us which part you don't understand so we can help you out!

Okay, so the main assumption here is that the base height is one metre ie. it finishes its projectile motion at an altitude of 1m. This simplifies things a lot

Subbing into the equation v=u+at, knowing that v is zero at the vertex and acceleration is -9.8 m/s/s, and also that time of flight to the vertex is half of 6.1, you can rearrange to get that:

And gets you an answer of 29.89m/s, or just 30m/s to one significant figure.

Using this information, sub it into s=ut+(1/2)at^2, using the fact that u=29.89, t=3.05, a=-9.8. The distance going to the vertex is 45.58225m, but since it comes back down we double it to get 91.1645m. To two significant figures its just 91m.

Hope this right and hope this helps

Hi Jirachi,
Thank you for replying,
That's what I did at first and also obtained the same result, but something doesn't sit right.
Why was the "1m" given in the information?
My interpretation of the question then becomes, say, a kid is holding a pinata at 1m , throws it up in the air, and it eventually lands on the ground (0m) in 6.1 seconds. Is it possible to solve this without the initial velocity?
I don't have the solution to this question.

[fun_jirachi]

Why was the "1m" given in the information?
My interpretation of the question then becomes, say, a kid is holding a pinata at 1m , throws it up in the air, and it eventually lands on the ground (0m) in 6.1 seconds. Is it possible to solve this without the initial velocity?
I don't have the solution to this question.

This is definitely possible and that's something I had trouble getting around, hence the assumption. I'll edit in the solution with those assumptions in mind later

EDIT: Okay, this is what I did with this new angle of attacking the question

All you really know is the time of flight, 6.1, and that the displacement to the vertex going up is one metre less than the displacement going down. Using this, you can let the time going up be x, and the time going down be 6.1-x. Similarly, you can let the displacement going up by s, and the displacement going down be s+1.

Knowing this, we need some way to link v and u before moving on to use v=u+at twice. Linking the two using the information deduced above gets you the relationship that v²=u²+19.6, where v is the final velocity of the pinata hitting the ground and u is the initial velocity of the pinata as it is projected upwards. Knowing this relationship now, and knowing also that v=0 going up, and u=0 going down, utilise v=u+at.

You should get to a line that looks somewhat like this (this is the v=u+at from going down) (the LHS from what we just did, the RHS from our initial deductions, linked by v=u+at):


From there, solve for u by using v=u+at going up, finding that u is equal to 9.8x. Notice that you don't actually care what x or u are at this point, it's just an in between sort of thing. Sub this into the u squared, then solve for x. Then sub your value for x into u=9.8x to find your value for u. I got 29.72m/s, or 30m/s to 1SF. It's super marginal, only 0.17m/s different than if you factored out the 1m.

From here, it's pretty obvious. Now you have your t value going up, which is roughly 3.033, I think you should be able to sub into s=ut+(1/2)at² twice and then add them up to get the total s. Don't forget to subtract one at the end. You should get 90m (2SF)

In the end, you do get Robot10's answer (good job bud) but like it's a hell of a lot more time consuming. As you can probably see, the time differential going up and down using this method is 0.03s, but the only noticeable difference it makes is to the answer in part b) which is 1 metre less. Make of it what you will