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Sagot :
Explanation:
Vf=Vi+at
reference frame,take down as positive in y- axis
Data
Vi=0m/s
Vf=?
a=g(9.8m/s2)
t=6s
Vf=0+9.8m/s2*6s
Vf=58.8m/s
The velocity of the object after falling for 6 s from a high building is 58.8 m/s
- From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Time (t) = 6 s
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) =?
- The velocity of the object after falling for 6 s can be obtained as follow:
v = u + gt
v = 0 + (9.8 × 6)
v = 0 + 58.8
v = 58.8 m/s
Therefore, the velocity of the object after falling for 6 s is 58.8 m/s
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