A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet.
Compared to the acceleration due to gravity near Earth’s surface, the acceleration due to gravity near the
surface of the planet is approximately
On the planet that is NOT earth,
Y (1/2)g'*t^2
22 = (g'/2)*(3.0)^2
so the value of the acceleration of gravity there is
g' = 44/9 = 4.89 m/s^2
The is just about half the value of g at the surface of the Earth (9.8 m/s^2)
To find the acceleration due to gravity near the surface of the planet, we can use the equation for free fall:
d = (1/2) * g * t^2
where:
- d is the distance (fallen)
- g is the acceleration due to gravity
- t is the time taken to fall
In this case, the distance fallen (d) is 22 meters and the time taken (t) is 3.0 seconds.
Rearranging the equation to solve for g:
g = (2 * d) / (t^2)
Substituting the given values:
g = (2 * 22) / (3.0^2)
g = (44) / (9.0)
g ≈ 4.89 m/s^2
Therefore, the acceleration due to gravity near the surface of the planet is approximately 4.89 m/s^2.