i need help with this question.?

Question

ok i need help in a physics problem

we launch a cannon into the air at 28 degrees (we neglect air resitance) at ground level.

I know all the info for when it is at the highest point (right before it drops)

the time is .296 seconds
Dx=1.64m
Dy=.43808m
Vx=5.56m/s
Vy= 2.96m/s

now i need to know Dx, Dy, Vx, Vy for these times
.0592
.1184
.1776
.2368
.3552
.4144
.4736
.5328
.592

i would really like to know how to do this (or the answers because i think im doing it wrong, i want to check) PLEASE HELP ME AND THANKYOU

Answer 1

Your question is not clear. The parameters you list, Dx,Dy,Vx,Vy supposedly occur at the top of the trajectory, which is at t = .296 sec. However, at the top of the trajectory, the vertical component of velocity is zero.

Is this supposed to happen on earth with accel of gravity = 9.8m/sec^2? This does not square with the numbers you gave. Vertical velocity Vy = Vy0 - g*t, (Vy0 is the initial velocity at ground level); vertical displacement Dy = Vy0*t - .5*g*t^2. From the first, Vy0 = Vy + g*t. Using t = .296, get Vy0 = 5.86m/sec Use that with t in the second to get Dy = 1.3m. So I'm not sure what is happening here; I'm probably not interpreting your question properly.

Answer 2

One short cut I like to use for these things is propotions. Okay. Lets see here (I might screw you up more, so the best source is your teacher still)....and with the minimum vague details you have given...

When time= .296, Dx=1.64

therefore, when time= 0.0592

0.0592/0.0296 is 1.64/x

solve for x and get Dx when time is 0.0592.

Since you know all the info, for all Distance and velocity at initial time, I don't see the importance of 28 degrees for the rest of the equation.

Good luck (and if what I just did is right, I need to get my butt back to my cal class ;))

Answer 3

Goodness gracious, the best thing to do is talk to your teacher...can't be helped here.

Answer 4

Holy ****. I am so confused. Why the hell do we have to do this stuff.?!