Question

A stone of mass m is dropped from a sheer cliff of height h. The wind exerts a force F on the stone. Assume the wind is parallel to the face of the cliff and the ground and F is constant. (Use any variable or symbol stated above along with the following as necessary: g. For all vectors, enter the magnitude.) If t = 0 corresponds to the moment the stone is released, at what time t does it strike the ground?

Answer 1

The stone will strike the ground in approximately 1.08 seconds.

Step-by-step explanation:

The time it takes for a stone to strike the ground can be determined using the equations of motion. We can use the equation for vertical position: h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity, and t is the time. Since the stone is dropped and there is no initial velocity in the vertical direction, the equation simplifies to h = (1/2)gt^2.

Given that the stone is dropped from a height h and the equations assume downward as negative, we can rewrite the equation as h = -gt^2/2. Rearranging the equation to solve for t, we get t = sqrt(2h/g).

Substituting the given values into the equation, we have t = sqrt(2(6 m)/9.8 m/s^2) ≈ 1.08 s. Therefore, it will take approximately 1.08 seconds for the stone to strike the ground.