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?
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.
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
Based on your calculation, it seems like you correctly identified the formula to use, which is the conservation of mechanical energy equation: gH = Δ(v^2)/2, where g is the acceleration due to gravity, H is the height of the incline, and Δ(v^2) is the change in speed squared.
However, there is an error in plugging in the numbers. Let's correct it step by step:
First, find Δ(v^2) by subtracting the initial speed squared from the final speed squared:
Δ(v^2) = (29.7^2) - (6.93^2) = 880.09 - 48.02 = 832.07 m^2/s^2
Next, divide Δ(v^2) by 2 and divide the result by g to find the height H:
H = (Δ(v^2))/ (2 * g) = 832.07 / (2 * 9.8) = 42.45 m
Therefore, the length of the incline is approximately 42.45 meters, not 40.45 meters.
Make sure to double-check your calculations to ensure accuracy.