In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills.) To demonstrate this, find the final speed in m/s and the time taken in seconds for a skier who skies 69.0 m along a 26° slope neglecting friction for the following two cases. (Enter the final speeds to at least one decimal place.) (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events?

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!