The question is posted here:
http://www.flickr.com/photos/69230722@N0…
I have done part (a) through the use of SUVAT equations, and got the answer to be 9.375m/s exactly (which the book says is correct)
However, I do not know how to do part b or c.
(This is probably not helped by the fact that I don't really understand what it means when it says "his centre of gravity falls 0.95m between his take-off and landing").
If you could explain how to do those last 2 parts of the question that would be brilliant!
Thank you!
P.S. The book says the answer to part (c) is 16 degrees
http://www.flickr.com/photos/69230722@N0…
I have done part (a) through the use of SUVAT equations, and got the answer to be 9.375m/s exactly (which the book says is correct)
However, I do not know how to do part b or c.
(This is probably not helped by the fact that I don't really understand what it means when it says "his centre of gravity falls 0.95m between his take-off and landing").
If you could explain how to do those last 2 parts of the question that would be brilliant!
Thank you!
P.S. The book says the answer to part (c) is 16 degrees
-
they're just telling u if u treat the athlete as a particle its vert displ=-0.95m during jump
OK using S=Vt+at^2/2..wotchout for signs as well as sines..
>-0.95=0.8*V-4.9*0.64 (g=-9.8,t=0.8s, V=vert comp of vel)
Solving>V=2.7325m/s=vsin(thet)
Ist bit vcos(thet)=horiz comp=9.375m/s
dividing horiz/vert..>thet=arctan(0.2915)=16.24de…
OK using S=Vt+at^2/2..wotchout for signs as well as sines..
>-0.95=0.8*V-4.9*0.64 (g=-9.8,t=0.8s, V=vert comp of vel)
Solving>V=2.7325m/s=vsin(thet)
Ist bit vcos(thet)=horiz comp=9.375m/s
dividing horiz/vert..>thet=arctan(0.2915)=16.24de…