Question

a worker aciddentally kicks a rock off of a roof of a 5m tall building at 2 m/s. what is the distance (range) x from the edge of the building where the rock will land?

Answer 1

Answer:
We can use the equations of motion to find the horizontal distance (range) x from the edge of the building where the rock will land. The vertical motion of the rock is governed by the acceleration due to gravity, which is constant and equal to 9.81 m/s^2.

First, we can find the time it takes for the rock to hit the ground using the vertical motion equation:

• y = vi*t + (1/2)at^2

where y is the initial height of the rock (5 m), vi is the initial vertical velocity of the rock (0 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time it takes for the rock to hit the ground.

Solving for t, we get:

• t = sqrt((2*y)/a) = sqrt((2*5)/9.81) = 1.02 s

Now we can find the horizontal distance (range) x from the edge of the building using the horizontal motion equation:

• x = vx*t

where vx is the horizontal velocity of the rock, which is constant and equal to 2 m/s.

Substituting the values we have, we get:

• x = 2 m/s * 1.02 s = 2.04 m

Therefore, the distance (range) x from the edge of the building where the rock will land is 2.04 meters.