Question
A skier is gliding along at 6.93 m/s on horizontal, frictionless snow. He suddenly starts down a 17.2° incline. His speed at the bottom is 29.7 m/s. What is the length of the incline?
i understand that the equation used is gH= change in (v^2)/2
so i plug in
9.8H=(29.7^2-6.93^2)/2
9.8H=396.5
H=396.5/9.8
H=40.45 m
but this answer still comes up as wrong? can anyone help me with where i'm going wrong?
i understand that the equation used is gH= change in (v^2)/2
so i plug in
9.8H=(29.7^2-6.93^2)/2
9.8H=396.5
H=396.5/9.8
H=40.45 m
but this answer still comes up as wrong? can anyone help me with where i'm going wrong?
Answers
bobpursley
YOu found the vertical drop on the incline. The distance is quite something else, along the incline
H=dSin17.2
solve for d, given H above. I didn't check your math.
H=dSin17.2
solve for d, given H above. I didn't check your math.
Jason
so thatd be H=dSin17.2
40.45m=dsin17.2
d=40.45/sin17.2
d=136.8
it all looks right, but for some reason incorrect, if anyone has a chance, can you check my math for me? i cant understand where i went wrong
40.45m=dsin17.2
d=40.45/sin17.2
d=136.8
it all looks right, but for some reason incorrect, if anyone has a chance, can you check my math for me? i cant understand where i went wrong