(Based on Knight, Physics for Scientists and Engineers) You’ve taken a summer jo

Question

(Based on Knight, Physics for Scientists and Engineers) You’ve taken a summer job at a water park. In one stunt, a water skier is going to glide up a 2.0-m-high, 10.0 m long ramp, then sail through the air over a 5.0-m-wide tank filled with hungry sharks. The coefficient of kinetic friction between the skis and the ramp is 0.15. You will be driving the boat that pulls her to the ramp. She’ll drop the tow rope at the base of the ramp just as you veer away. What is the minimum speed must you have as you reach the ramp in order for her to live to do this again tomorrow?

Explanation / Answer

height of ramp, h = 2 m
length of ramp, l = 10 m
tank width, w = 5 m
coefficneit of kinetic fricdeion, k = 0.15
let the minimum speed before reaching the ramp be u
then speed at the edge of ramp = v
let angle of ramp be theta
then vcos(theta)*t = w
and, vsin(theta)*t - 0.5gt^2 = h
vsin(theta)*w/vcos(theta) - 0.5gw^2/v^2cos^2(theta) = h
tan(theta)*w - 0.5gw^2/v^2cos^2(theta) = h

now, sin(theta) = 2/10
theta = 11.54 deg

so time taken to fall off the ramp = t
so, vsin(11.54)*t - 0.5*9.81*t^2 = -2
4.905t^2 - 0.2vt - 2 = 0
v = 5/cos(11.54)*t
4.905t^2 - 0.2*5/cos(11.54) - 2 = 0
t = 0.6158 s
hence v = 8.2859 m/s

now, for mass of person = m
work done by friction = k*mgcos(theta)*L
so, from conservation of energy
0.5mu^2 = 0.5mv^2 + mgh + k*mgcos(theta)*L
0.5u^2 = 0.5v^2 + 9.81*2 + 0.15*9.81*cos(11.54)*10
u = 11.693 m/s

so minimum initial speed must be 11.693 m/s

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