Respuesta :
s = ut + 0.5at^2
10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414}
t^2 ~ 2.04
t ~ 1.43 seconds
10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414}
t^2 ~ 2.04
t ~ 1.43 seconds
Given the height from which the marble was released from and the effect of gravity, it will take approximately 1.43 seconds for the marble to hit the ground.
Given the data in the question;
The marble was initially at rest before it was released,
- Initial velocity; [tex]u = 0m/s[/tex]
- Height or distance from which it was released from; [tex]s = 10m[/tex]
We determine the time taken for the marble to hit the ground.
From the Second Equation of Motion:
[tex]s = ut + \frac{1}{2} gt^2[/tex]
Where s is the height or distance, u is the initial velocity, t is the time taken and g is acceleration due to gravity( [tex]9.8m/s^2[/tex] )
We substitute our values into the equation
[tex]10m = 0 + \frac{1}{2}\ *\ 9.8m/s^2\ *\ t^2\\\\10m = 4.9m/s^2\ *\ t^2 \\\\t = \sqrt{\frac{10m}{4.9m/s} }\\\\t = 1.43s[/tex]
Therefore, given the height from which the marble was released from and the effect of gravity, it will take approximately 1.43 seconds for the marble to hit the ground.
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