Respuesta :
Okay assuming the log had no initial vertical motion (you threw it straight out), the distance to the bottom is
dist = 1/2 gt^2 = 1/2 x 9.8m/s/s x
(5.65s)^2 = 156.4m
the velocity at the bottom = g t = 9.m/s/s x 5.65s =55.4m/s
Hope this helps!
dist = 1/2 gt^2 = 1/2 x 9.8m/s/s x
(5.65s)^2 = 156.4m
the velocity at the bottom = g t = 9.m/s/s x 5.65s =55.4m/s
Hope this helps!
Answer:
55.4265 m/s
513.71 ft
Explanation:
t = Time taken to hit the ground = 5.65 seconds
a = Acceleration due to gravity = 9.81 m/s²
u = Initial velocity = 0 m/s
v = Final velocity
From equation of motion
[tex]v=u+at\\\Rightarrow v=0+9.81\times 5.65\\\Rightarrow v=55.4265\ m/s[/tex]
∴ Velocity of the log at impact is 55.4265 m/s
Distance
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 5.65+\frac{1}{2}\times 9.81\times 5.65^2\\\Rightarrow s=156.5798625\ m[/tex]
Converting to feet
1 m = 3.28084 ft
156.5798625 m = 156.5798625×3.28084 ft
= 513.71 ft
∴ The distance the log fell is 513.71 ft
