Kinematics: Finding Acceleration Due to Gravity on a Planet with No Atmosphere

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To find the acceleration due to gravity on a planet with no atmosphere, the kinematic equation D = v0t + 1/2 * g * t^2 is applicable. Given that the object falls 2.66 meters in the first second from rest, the initial velocity (v0) is zero. By substituting the known values into the equation, the acceleration (g) can be calculated. This approach effectively utilizes the provided time, initial velocity, and distance to solve for gravitational acceleration. The final result will be expressed in units of m/s^2.
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An object is released from rest on a planet that has no atmosphere. The object falls freely for 2.66m in the first second. What is the magnitude of the acceleration due to gravity on the planet? Answer in units of m/s^2

What kinematic equation would I use for this?
 
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D=v0t + 1/2 *g*t2 might be a good place to start.

You have time, initial velocity, and distance. So solve for acceleration (g)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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