Projectile motion (ground to ground)

Click For Summary
The discussion revolves around solving a projectile motion problem involving a baseball hit at a height of 1.30 m and caught at the same height after traveling a horizontal distance of 42 m alongside a wall. Key calculations include determining the horizontal distance traveled, the magnitude and angle of the ball's velocity just after being hit, and the height of the wall. The average horizontal velocity while the ball is over the wall is calculated to be 10.77 m/s, with discussions on using kinematic equations for further analysis. Participants also share tips on using LaTeX for typing unit vectors with hats. The conversation emphasizes the importance of understanding projectile motion principles and accurate mathematical representation.
J-dizzal
Messages
394
Reaction score
6

Homework Statement


In the figure, a baseball is hit at a height h = 1.30 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.1 s after it is hit and then down past the top of the wall 3.9 s later, at distance D = 42 m farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?
http://edugen.wileyplus.com/edugen/...u/ch0/EAT_13496164302600_3698173899564502.GIF

Homework Equations


I have the kinematic equations and equations for motion (while in flight)

The Attempt at a Solution


known solns:
ay = -g = -9.8m/s2
v0y = 29.89 m/s
vfy = -29.89 m/s
Δy = 0
t= 6.1s

im stuck at this point. i think maybe i can set the v at max height equal to the x component of v0?[/B]
 
Physics news on Phys.org
Consider its horizontal velocity for the time it is above the wall.
 
haruspex said:
Consider its horizontal velocity for the time it is above the wall.
velocity for the time the ball is over the wall,
v = Δd/Δt
= 42m / 3.9
= 10.77m/s.
But this is average velocity, if ax is constant and vx is constant then the avg velocity is equal to v at any point?

Then,
vavg = v0x = v0cosθi,
is the x component of v0.

Thanks.
How do you type unit vecotors i,j, and k with hats?
 
J-dizzal said:
velocity for the time the ball is over the wall,
v = Δd/Δt
= 42m / 3.9
= 10.77m/s.
But this is average velocity, if ax is constant and vx is constant then the avg velocity is equal to v at any point?

Then,
vavg = v0x = v0cosθi,
is the x component of v0.

Thanks.
How do you type unit vecotors i,j, and k with hats?
Yes.
For the hat, use Latex. Putting a consecutive pairs of hash symbols(#) starts and ends a Latex string. Using dollar symbols instead puts the Latex on a line of its own. A hat over a character is achieved by writing \hat in front of it. Put a space after hat. If it is to go over something more complicated than a single character, use curly brackets: \hat{x^2} will produce ##\hat{x^2}##.
You can see how others have written latex by right-clicking on the generated text in their post and selecting View math as. You can then cut and paste to produce your own.
Sometimes the Latex refuses to work properly even though what you've written looks right. It's usually because you've included an unusual ASCII character or have some font change within it. Latex is jealous - in the Latex string, all must be Latex.
Always use the Preview button to check the Latex before posting.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
Replies
40
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
16
Views
3K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
4K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K