Answer:
9.4 m/s
Step-by-step explanation:
We are given that
Height of pile driver above a pile =h=4.5 m
Acceleration due to gravity=[tex]g=9.8m/s^2[/tex]
According to law of conservation of energy
K.E=Potential energy
[tex]\frac{1}{2}mv^2=mgh[/tex]
[tex]v^2=2gh[/tex]
[tex]v=\sqrt{2gh}[/tex]
Substitute the values
[tex]v=\sqrt{2\times 9.8\times 4.5}[/tex]m/s
[tex]v=9.4 m/s[/tex]
Hence, the velocity of pile driver as it hits the pile=9.4 m/s
The pile driver falls under the acceleration due gravity, such that the
velocity at the pile is √(2·g·h).
Response:
The given parameter are;
Height from which the pile driver falls = 4.50 m
Required:
The velocity of the pile driver as it hits the pile.
Solution:
The equation of the velocity of the object in free fall is given as follows;
v² = u² + 2·g·h
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
u = The initial velocity = 0
h = The height = 4.50 m
Which gives;
v² = 0 + 2·g·h
v = √(2·g·h)
v = √(2 × 9.81 × 4.50) ≈ 9.396
The velocity of the pile driver when it hits the pile, v ≈ 9.396 m/s
Learn more about free fall motion here:
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