okay, i think i got the ansewr.
So i used vf^2 = vi^2 + 2a(yf-yi)
vf2 = (8.49)^2 + 2 (-9.81(0-1.72))
vf = 10.29
so,
resultant = ((10.29)^2 + (14.71)^2)^0.5
= 17.95 so this is the final velocity (?)
then total momentum would be
(2kg * 16.99m/s) + (2kg * 17. 95)
= 69.88
is...
well the questions are:
a) what is the distance of the throw
b) what is the total momentum of the discus on landing?
c) create a type 1 scatterplot to show the relationship between the
projection angles, 0 degrees to 90 degrees, and the throw distance
d) what is the optimal angle of...
hmm I'm sooo lost.
initial speed is 16.99.
direction and height?
and how do i find the vertical and horizontal motion?
i know that the initial vertical velocity is 8.49 (16.99sin30)
and horizontal velocity is 14.71 (16.99cos30)
what do i do after that?
this is what I've done so...
Hmmm I don't understand how PE and KE comes into this...
I practically know nothing about physics so...haha.
Well i know that the initial momentum is 33.98 because i just
multiplied the weight of the discus (2kg) times the velocity
it was going (16.99m/s). But then, how do i figure out the...
Hi, I'm new here. I have a question with one problem and i was wondering if any of you could help me out!
So um here's the problem..
A high school discus thrower rotates at an average angular acceleration of 895 degrees/sec^2 for 1.25 sec. The 2kg discus is realeased 0.87 meters away from...