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A 50 kg bobsled slides down an ice track
starting (at zero initial speed) from the top of
a(n) 173 m high hill.
The acceleration of gravity is 9.8 m/s^2.
Neglect friction and air resistance and determine the bobsled’s speed at the bottom of
the hill.
Answer in units of m/s.


Sagot :

Hi there!

We can use the following kinematic equation:

[tex]v_f^2 = v_i^2 + 2ad[/tex]

The initial velocity is 0 m/s, so:

[tex]v_f^2 = 2ad[/tex]

vf = final velocity (? m/s)
a = acceleration due to gravity (g)
d = vertical height (m)

Plug in the givens and solve:

[tex]v_f = \sqrt{2gd} = \sqrt{2(9.8)(173)} = \boxed{58.23 \frac{m}{s}}[/tex]

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