Physics, 24.12.2019 19:31 makinzy03

)a skier starts down a frictionless 32° slope. after a vertical drop of 25 m, the slope temporarily levels out and then slopes down at 20°, dropping an additional 38 m vertically before leveling out again. find the skier’s speed on the two level stretches

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Answer from: karyvillanuevap5w9ym

The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.

Explanation:

Given that,

Slope down = 32°

Height = 25 m

Before leveling,

Slope down = 20°

Height = 38 m

We need to calculate the skier’s speed on the two level stretches

Using formula of energy

P.E=K.E

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

For first stretch,

Height = 25 m

Put the value into the formula

$v=\sqrt{2\times9.8\times25}$

$v=22.13\ m/s$

For second stretch,

Height = 38 m

Put the value into the formula

$v=\sqrt{2\times9.8\times38}$

$v=27.29\ m/s$

Hence, The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.

Answer from: Quest

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

Answer from: Quest

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