Question

A 0.145 kg baseball is thrown at launch angle of 30° and strikes the ground at 18 m/s. How fast would it be moving when it reaches the ground if its launch angle were 45°? Ignore air resistance.

Answer 1

You should pull out the projectile motion equations for this haha. I'll see what I can do when I get home.

Answer 2

Oh gosh, okay

Answer 3

When will you get home?

Answer 4

About an hour from now. But get the equations ready in the post xD

Answer 5

Oh dear XD I will! Hopefully I'll still be on in an hour hahaha

Answer 6

I think I just remembered how to do this one! Not so much an equation, but just a few steps I need to take... Thank you for your help with the other problem haha! I think I'm good now :)

Answer 7

Oh alright. Good job :)

Answer 8

Thank you :)

Answer 9

No problem.

Answer 10

I was wrong XD I'm not going to be available in an hour, so can someone else help?

Answer 11

@roadjester ? Can you help?

Answer 12

or @Mashy ?

Answer 13

give me a sec to finish up, unless @Mashy you want to take this one?

Answer 14

ermm.. ll try

i ll give you a clue,
the velocity with which any object hits the ground is the same velocity with which it was launched (as long as hit point and the launch point are at the same level)

Answer 15

I'm trying to figure out what formula that fits in.... Hmmm This is a hard one

Answer 16

try to use vectors and decompose the velocities..

is that what you have been taught in school? is that how you deal with projectile motions?

Answer 17

This is actually the energy unit... I don't know how to decompose the velocities :/

Answer 18

So it should fall into the mechanical energy set of formulas

Answer 19

no no.. this is from kinematics.. 2d kinematics!

Answer 20

That's what I thought when I first saw it too! I have no idea why it's in my energy worksheet :/

Answer 21

lol.. thats ok.. now can you try and work this out?

Answer 22

No... I'm still not positive kinematics will help with this :(

Answer 23

What do you think @roadjester ? This question is beginning to depress me :(

Answer 24

Is the velocity again 18m/s?

Answer 25

Yeah

Answer 26

@roadjester u do this ok :D thanks!

Answer 27

Wait... I kinda remember something about the angle not impacting the velocity? If the distance doesn't change?

Answer 28

Maybe I am making that up :/

Answer 29

nope. This is a trick question. Kinda.
The angle does not affect the final velocity.

Answer 30

So the answer is 18 m/s

Answer 31

uh huh

Answer 32

however I do not know the exact mechanics behind this (forgive the pun)
@ybarrap could you explain why the final velocity remains the same?

Answer 33

Any chance we can move on to my next (and final!) question while ybarrap answers? I really want to know why that is the case, but I also really want to be done with this worksheet.

Answer 34

idk

Answer 35

I can close this topic and open a new one... ybarrap can still answer if it's closed, right? I really want to understand this problem :)!

Answer 36

@roadjester is correct. The velocity at the end is the same as the velocity at the beginning.

Trajectories follow the path of a parabola. So at the beginning, the projectile will have a certain velocity, it will reach a minimum speed at the maximum height of the parabola, which is at the center of the parabola (looking at the x-axis) where it will then begin to descend, following a path that is a mirror-image to that which it just experienced. The mirror image is not only in shape (i.e. 2nd half of parabola) but also in speed (but negative).

So if we start at the top of the parabola, where the speed is zero, it will begin to descend. At each height (h) from the ground, the velocity will be exactly the same as it was on the other side of the parabola (opposite center), where it was going up. By symmetry, we then conclude that the velocity when it reaches the ground at the end is exactly the same as when it was launched, but just negative in sign: