Catapult
A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 33.0 m above sea level, directed at an angle theta = 45.9° above the horizontal, and with a speed v = 28.2 m/s.
Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.
1.06×102 m
Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.
1.06×102 m
Discussion
Anonymous 1 (Thu Jan 21 07:48:22 pm 2010 (EST))
How do we do this problem? HELP Please
NEW
Anonymous 1 (Sat Jan 23 01:28:15 pm 2010 (EST))
Can anyone please help me on this one!!!
NEW
Hi, I am having a real hard time trying to figure this out. I've done the quradratic
equation and I figured out the plus and minus results, but I don't know what to
do next!
Can anyone help me... PLEASE!!!!!!
NEW
Re: Please HELP!!!! Anonymous 3 (Sat Jan 23 09:36:44 pm 2010 (EST))
Here's what I did.
First I used the equation y=yo+vo*(sin theta)*t + .5gt^2
Everything you need will be there except for time. Rearrange
that such that you'll have an equation that can be solved
using the quadratic equation (there's that again), and use
the bigger value of t for time.
Plug that into x = vo* (cos theta)* t , which you just
found, and there you have it.
Just to clarify, yo is the initial height, and vo is the
initial velocity.
NEW
Right now I just cannot understand why I'm not getting this right...I got 3.555
and .1225 for the time and when I plug it into that equation I have got 3 wrong
answers....can anyone tell me if my time's are right and why I'm not getting this.
NEW
Anonymous 5 (Sun Jan 24 02:55:03 pm 2010 (EST))
make sure that you have g as a negative value.. thats why your time values are
wrong... you should get only one positive answer out of the quadratic equation
and use that value to plug into the equation he explained above...
NEW
A little more help please Anonymous 6 (Sun Jan 24 05:39:19 pm 2010 (EST))
Am I using the correct A,B,and C for the quadratic equation?
A=Height(35.0)
B=Vo(30.8)
C=sin(theta)(47.3deg)
NEW
Thank you so much, I got it!!! :D
How do we do this problem? HELP Please
NEW
Anonymous 1 (Sat Jan 23 01:28:15 pm 2010 (EST))
Can anyone please help me on this one!!!
NEW
Hi, I am having a real hard time trying to figure this out. I've done the quradratic
equation and I figured out the plus and minus results, but I don't know what to
do next!
Can anyone help me... PLEASE!!!!!!
NEW
Re: Please HELP!!!! Anonymous 3 (Sat Jan 23 09:36:44 pm 2010 (EST))
Here's what I did.
First I used the equation y=yo+vo*(sin theta)*t + .5gt^2
Everything you need will be there except for time. Rearrange
that such that you'll have an equation that can be solved
using the quadratic equation (there's that again), and use
the bigger value of t for time.
Plug that into x = vo* (cos theta)* t , which you just
found, and there you have it.
Just to clarify, yo is the initial height, and vo is the
initial velocity.
NEW
Right now I just cannot understand why I'm not getting this right...I got 3.555
and .1225 for the time and when I plug it into that equation I have got 3 wrong
answers....can anyone tell me if my time's are right and why I'm not getting this.
NEW
Anonymous 5 (Sun Jan 24 02:55:03 pm 2010 (EST))
make sure that you have g as a negative value.. thats why your time values are
wrong... you should get only one positive answer out of the quadratic equation
and use that value to plug into the equation he explained above...
NEW
A little more help please Anonymous 6 (Sun Jan 24 05:39:19 pm 2010 (EST))
Am I using the correct A,B,and C for the quadratic equation?
A=Height(35.0)
B=Vo(30.8)
C=sin(theta)(47.3deg)
NEW
Thank you so much, I got it!!! :D