Answer:
Vf = 73.4 m/s
Explanation:
This is the case of vertical motion where we have to find the final velocity of the penny when it hits the ground. We can use 3rd equation of motion to find the final velocity:
2gh = Vf² - Vi²
where,
g = 9.8 m/s²
h = height = 275 m
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s
Therefore,
2(9.8 m/s²)(275 m) = Vf² - (0 m/s)²
Vf = √5390 m²/s²
Vf = 73.4 m/s
The speed with penny dropped from the Golden Gate Bridge hits the ground is 73.4m/s.
Given the data in the question;
Since the penny was at rest before it was dropped
Final velocity; [tex]v = \ ?[/tex]
To determine the speed with which the penny hits the ground, we use the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, s is the height of the bridge and a is acceleration due to gravity ( since its under gravity; [tex]a = g = 9.8m/s^2[/tex] )
We substitute our values into the equation
[tex]v^2 = 0 + [ 2\ *\ 9.8m/s^2\ *\ 275m]\\\\v^2 = 5390m^2/s^2\\\\v = \sqrt{5390m^2/s^2}\\\\v = 73.4 m/s[/tex]
Therefore, the speed with penny dropped from the Golden Gate Bridge hits the ground is 73.4m/s.
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