Recent content by monac

In the back of the book, the answer for d was x ~ 0.405 m

So imagine that vi = 0?

Oh the whole thing is under the square root. t = (vi sin35 + √(vi^2 sin^2 35 + 23.544)) / 9.81

it's a long process. I ll just type the main parts of the answers. (a) x = vicos35t t = (vi sin35 + √vi^2 sin^2 35 + 23.544) / 9.81 so x = vi cos35 ((vi sin35 + √vi^2 sin^2 35 + 23.544) / 9.81) for b and c i just had to plug in values for vi and get the x so that was easy. I have...

A spring cannon is located at the edge of a table that is 1.2 m above the floor. A steel ball is launched from the cannon with speed Vo at 35.0 degrees above the horizontal. (a) Find the horizontal displacement component of the ball tothe point where it lands on the floor as a function of Vo...

Thank you! :)

so if I solved the integral and I got like r(t) = 8t^2 + 5t I put the arrow on top of the arrow since it's a vector. I get that ... But do I still include the i and j? so would it be r(t) = 8t^2 i + 5t j ?

So it says that a particle is going with a velocity of 5i m/s at t = 0 and varying acceleration a = 6√t j. it asked me to find the velocity and position of the particle as a function of time. So i did an integral of the acceleration to get velocity and did the integral of the velocity to get...

I am trying to find a ... what am I doing with the s ? I am so close to giving up on this problem

how do I get the s when I don't know the t?

your process is still giving me 4.07 m/s^2 which is the wrong answer :(

You no longer have the absolute value symbols where? in the second part of my solution?

so 2s/u+v t 2*23/7.9+15.8 = t 1.94 s = t vf - vi / t = a 15.8 - 7.9 / 1.94 = a a = 4.07 m/s^2 <-- That's the wrong answer I got in the first place :( This is disappointing.

So I understand that our initial velocity should be positive, then ... what equation do I plug it into? all equations have a t in them ... which is unknown.

so you are saying v0 = 0 m/s vf = 15.8 m/s ? Then what would be delta x? 15.8^2 - 0^2 / 2*23? Do you think that's right?