3rd week of beginner physics, brain fried

[picklepie159]

1. Ok, hi everyone, I'm new to the forum. I'm hoping you can help me a bit because I'm in trouble now, and a few concepts need clearing up.
Ok, so there are two questions
#1 A ball is rolled off a table 1 meter high, and lands 4 meters away. what is the velocity for the x vector? What is the angle at which it hits the floor?

#2 A (hyperelastic? superelastic? perfectly elastic? I forgot) ball is rolled down the edge of a well at a horizontal speed of 4 m/s. The well is 4 meters deep and 0.5 meters across. What is the number of times the ball hits the wall?

Homework Equations

What I was thinking of was these...
1 meter = 1/2g t^2 To find the time
4 meters= velocity x (t)
I am probably wrong, but to find the angle at which the ball lands at, I will have to get the y-vector/x-vector = inverse tan. Is that correct? If so, then what should I put for the magnitude for the y vector? Should I put the final velocity, or average velocity on the y-axis?


For #2, I was planning to find the amount of time that it takes for a 4 m/s ball to cross 0,5 meters= ie. 0.5 m = 4 m/s T

Then, I would use D= 1/2 g t^2 to find how much it would drop for the first bounce. After the first bounce, with the velocity accelerating, how would i find out the amount of bounces for the rest of the distances?



3. The Attempt at a Solution - sort of explained above. Please help me in simple terms- I am but a mere 9th grader. Thanks for all the replies!

 

[zgozvrm]

What I was thinking of was these...
1 meter = 1/2g t^2 To find the time
4 meters= velocity x (t)

Looks good so far...
(depending on what you mean by "velocity")

... to find the angle at which the ball lands at, I will have to get the y-vector/x-vector = inverse tan. Is that correct?

Yes (kind of).
Actually y-vector/x-vector = tan(angle), so you'll need to take the inverse tan of y-vector/x-vector to get the angle.

... then what should I put for the magnitude for the y vector? Should I put the final velocity, or average velocity on the y-axis?

Are you looking for the initial y-vector, or the final y-vector?

 

[picklepie159]

okay, maybe not magnitude, but then how would i be able to find out what the y vector's supposed to be? Now I'm confused about the question itself :(

 

[picklepie159]

Ok, just asked my uncle over the phone, and he told me that to get the angle, it was simply
1meter/4meter= inverse tan. Is that it? Just wondering

 

[pat666]

No, what is the trajectory the ball would take? It's not a straight line which is what your uncle is thinking.
Find the final velocity in the x and y components then in that instant you can use trig.

 

[picklepie159]

wait- so with a time of 0.45 seconds, X velocity of 8.8 m/s, and a final y velocity of 0.441 meters, that would make out to be around 2.862 degrees? Is that right? Hrrmmm...

 

[pat666]

hopefully you know this bu Ill say it anyway, the units of velocity are m/s with some direction not metres.
known:
Sx=4 and Sy=1 S is displacement
a_y=9.81 and ax=0
uy=0 and ux=? u is the initial velocity
S=ut+1/2at^2
4=ut and 1=1/2*9.81t^2 goes to t=0.45 seconds
so 4=u*.45 goes to u=8.86m/s in the x direction initial and final velocity (no acceleration)
v^2=u^2 + 2as v is final velocity
v^2=2*9.81*1
v=4.429m/s final velocity in the y direction

so now we know that the final velocity is 8.86x+4.429y
so tan(theta)=4.429/8.86
theta=63.43 degrees!

Hope you understand my procedure?

 

[picklepie159]

Yeah! Thanks alot- it was the final y velocity which messed me up. I tried to use
V(final)=V(initial) + gt, making it
V(f) = 0 + 9.8(0.45)= 4.41
Except that I accidently kept on putting 0.98 as g instead of 9.8, causing lots of confusion.

As for the other problem, I hope that the following procedure might make sense.
Since the ball doesn't lose speed, I would treat it as if it had rolled off a table 4 meters high, going at 4 m/s. The time would be around 0.89 seconds. I would then calculate displacement on the x-axis with D= VT, and that would be around 3.57 meters. 3.57 m divided by 0.5 meters is around 7 bounces, right? Maybe...

Thanks again!

 

[pat666]

Sorry, I am not sure about your second question, you might have to wait for someone else or go see your teacher/professor.

 

[zgozvrm]

so now we know that the final velocity is 8.86x+4.429y
so tan(theta)=4.429/8.86
theta=63.43 degrees!

How did you come up with this?

4.429/8.86 [itex]\approx[/tex] 0.49989(without rounding off values for Vx and Vy, this actually comes out to exactly 0.5)

What is [itex]tan^{-1}(0.5)[/tex]? (It's not 63.43 degrees!)

 

[pat666]

Whooops, I must have put it in my calc upside down! (ie tan^-1(8.86/4.429)
Sorry.