tgilli said:Homework Statement
A ball is dropped from 1.5 y(m) and hits the ground in .54 t(s). What is the value of g (m/s^2)?
The accepted value of g = 9.8m/s^2
Homework Equations
y=1/2gt^2
The Attempt at a Solution
I'm getting this wrong, but this is my attempt.
1.5 = 1/2g x .54^2
1.5 = 1/2 x .292
1.5 = .146
.146 / 1.5 =
10.27 g (m/s^2)
10.27 g (m/s^2) is close, but not correct. The correct answer is 10.29 m/s^2
Clarification would be appreciated.
LowlyPion said:Calculate the expression with more precision. I get 10.288
Free fall is the motion of an object under the influence of gravity, without any other external forces acting on it.
Gravitational acceleration is the acceleration experienced by an object due to the force of gravity. It is a constant value and is approximately 9.8 m/s² on Earth.
The acceleration of an object in free fall is equal to the acceleration due to gravity. This means that the speed of the object increases by 9.8 m/s² every second.
No, the gravitational acceleration of an object does not change unless the object is moved to a different location with a different gravitational pull. The value of gravitational acceleration is constant for a given location.
The mass of an object does not affect its free fall or gravitational acceleration. All objects, regardless of their mass, experience the same acceleration due to gravity when in free fall.