Answer:
24.5
Step-by-step explanation:
Plz MaRk me as BrAiNliSt
Given:
Time to reach the ground, [tex]t = 4s[/tex]
Initial Velocity, [tex]u = 0[/tex] [stationary before dropped]
Acceleration, [tex]a = g = 9.8 \, ms^{-1}[/tex]
To find:
Height of the building, [tex]s[/tex]
Using the second equation of motion,
[tex]s = ut \, + (1/2)at^{2}\\\\\implies s = 0\, + (1/2)(9.8)(4)^{2}\\\implies s = (9.8)(8)\\\implies \bf \underline {s = 78.4 \, m}[/tex]
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Answers & Comments
Answer:
24.5
Step-by-step explanation:
Plz MaRk me as BrAiNliSt
Verified answer
Given:
Time to reach the ground, [tex]t = 4s[/tex]
Initial Velocity, [tex]u = 0[/tex] [stationary before dropped]
Acceleration, [tex]a = g = 9.8 \, ms^{-1}[/tex]
To find:
Height of the building, [tex]s[/tex]
Using the second equation of motion,
[tex]s = ut \, + (1/2)at^{2}\\\\\implies s = 0\, + (1/2)(9.8)(4)^{2}\\\implies s = (9.8)(8)\\\implies \bf \underline {s = 78.4 \, m}[/tex]
Hope this helps! Please mark as brainliest!