Answer:
a) final speed = 32.11 m/s; time = 4.30 s
b) final speed = 32.21 m/s; time = 3.98 s
c) It is still advantageous since difference in small amount of time is critical to determine whether the skier get 1st or 2nd place
Explanation:
We need to first find the actual height of the slope using trigo
sin(26) = h /69
h = 69 sin(26) = 52.62 m
For the whole questions, we must utilize the conservation of energy
Energy before = energy after
K.Ebefore + P.Ebefore = K.Eafter + P.Eafter
or KE1 + PE1 = KE2 + PE2
a) start from rest
initial velocity = 0
K.E1 = 0 because K.Ebefore = 1/2mv^2 = (1/2)m(0)^2 = 0
PE2 = 0 because final height become 0 (reaching lowest altitude)
Therefore
PE1 = KE2
mg(h1) = (1/2)m(v2)^2
g(h1) = (1/2)(v2)^2
9.8 (52.62) = (1/2) (v2)^2
v2^2 = 2(9.8)(52.62)
= 1031.352
final speed, v2 = sqrt(1031.352) = 32.11 m/s
to find time, t use equation of motion
s = (1/2)(u+v)t
rearrange to get t = 2s/(u+v)
t = 2(69)/(0+32.11)
t = 4.30 s
b) initial speed 2.5m/s
KE1 = (1/2)m(2.5)^2
PE1 = m(9.8)(52.62)
KE2 = (1/2)m(v2)^2
PE2 = m(9.8)(0) = 0
(1/2)m(2.5)^2 + m(9.8)(52.62) = (1/2)m(v2)^2
Reduce m through factorisation. So we can remove all m.
(1/2)(2.5)^2 + (9.8)(52.62) = (1/2)(v2)^2
3.125 + 515.676 = (1/2)(v2)^2
Rearrange to find v2
(v2)^2 = 2(3.125 + 515.676) = 1037.602
final speed, v2 = sqrt(1037.602) = 32.21 m/s
time, t = 2s/(u+v)
t = 2(69)/(2.5+32.21)
t = 3.98 s
c) The result is quite surprising since the difference between the 2 conditions is only 0.32s!
Though the difference is small, we shouldn't neglect the fact that the starting speed give extra boost (though small) to time taken. This is especially important for competitive sport where small time different determine difference between 1st and 2nd place!
