The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha
Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
To be referenced, it would be true
Answer:
Plane will 741.6959 m apart after 1.7 hour
Explanation:
We have given time = 1.7 hr
So if we draw the vectors of a 2d graph we see that the difference in angles is = 
Speed of first plane = 730 m/h
So distance traveled by first plane = 730×1.7 = 1241 m
Speed of second plane = 590 m/hr
So distance traveled by second plane = 590×1.7 = 1003 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.
Using the law of cosine,
representing the distance between the planes, we see that:

r = 741.6959 m
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