Answer:
Explanation:
The acceleration of an object down a slope (neglecting friction, µ = 0) is:
a = g × sin θ
Where,
g is the acceleration due to gravity and θ is the angle of the slope.
a = (9.8 × sin (21.5º)
= 3.592 m/s²
Using equations of motion,
S = ut + 1/2at²
Since, u = 0,
S = 1/2at²
347 = 1/2 × (3.592)t²
t² = 193.21
= sqrt(193.21)
= 13.9 s.
The climber will take 12.39 s to get out of the Avalanche, up the slope of mountain.
Given data:
The gravitational acceleration is, [tex]g = 9.8 \;\rm m/s^{2}[/tex].
The height of Avalanche is, h = 347 m.
The slope of Avalanche is, [tex]\theta = 27.5^\circ[/tex].
Here in this problem, we can apply the second kinematic equation of motion to obtain the time taken to get out of the way as,
[tex]h = ut+ \dfrac{1}{2}at^{2}[/tex]
Here, a is the acceleration of an object down a slope, and its value is,
[tex]a = gsin \theta\\\\a = 9.8 \times sin27.5\\\\a=4.52 \;\rm m/s^{2}[/tex]
And u is the initial speed, considering a climber to at rest initially, its value is u = 0. And t is the required time.
Solving as,
[tex]347 = 0 \times t+ \dfrac{1}{2} \times 4.52 t^{2}\\\\347 = 0 + \dfrac{1}{2} \times 4.52 t^{2}\\\\t = \sqrt{\dfrac{2 \times 347}{4.52} } \\\\t=12.39 \;\rm s[/tex]
Thus, a climber will take 12.39 s to get out of the Avalanche, up the slope of mountain.
Learn more about the kinematic equations here:
https://brainly.com/question/18937561
