A brick is dropped from a 78 m high scaffold, then the velocity as the brick hits the pavement is [tex]v_{f}[/tex] = 39.09 m/s
Explanation:
The brick is dropped from a 78 m high scaffold, So the direction of its velocity is downward. The initial velocity of brick is zero when it is dropped from 78 m high scaffold and we need to find its final velocity when it hits the pavement. So Using 3rd equation of motion
[tex]v_{f}^{2}-v_{i}^{2} = 2gS[/tex]
Where [tex]v_{f}[/tex] is the final velocity of the brick, [tex]v_{i}[/tex] is its initial velocity, g is acceleration due to gravity and S is the height of scaffold
[tex]v_{f}[/tex] = √(2*g*S)
[tex]v_{f}[/tex] = √(2*9.8*78)
[tex]v_{f}[/tex] = 39.09 m/s
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