If you drop a rock from rest and it travels for 3.3 s, how far does it fall?

Answer 1

If there is no air resistance, it falls by

d = (1/2)gt² = (1/2)(9.8 m/s²)(3.3 s)² = ? m

Answer 2

Hi Jessica,

Since you posted two questions about objects that are falling from rest I'm going to focus a bit heavier on the fundamentals of free-fall.

Background:

Free-fall is when an object is falling or rising with only the force of gravity acting on it. For example, a ball thrown into the air, or a skydiver jumping out of a plane. When in free-fall the acceleration of the object is a = g = 9.81 m/s² toward the ground. Knowing this we are able to apply fundamental kinematic equations to solving the object's position and velocity at any time.

For motion with constant acceleration:

1.  x = x₀ + v₀*t + 1/2a*t²
2.  v = v₀ + a*t
3.  v² = v₀² + 2a(Δx)

x₀: Initial Position

v₀: Initial Velocity

a: Constant acceleration

Since we know a = -g = 9.81 m/s² (acceleration due to gravity is negative because it points downward) we can simplify the above equations.

For an object in free-fall:

1. x = x₀ + v₀*t - 1/2g*t²
2.  v = v₀ - g*t
3.  v² = v₀² - 2g(Δx)

Solving your current problem:

To solve your current problem we will use equation 1. I chose this equation because we are looking to solve change in position (Δx = x - x₀)

x = x₀ + v₀*t - 1/2g*t²

We know v₀ = 0 because the object was released "from rest" and we know t = 3.3 s.

Plugging this into our equation we get:

x - x₀ = 0*t - 1/2*9.81*3.3²

x - x₀ = - 1/2*9.81*3.3² = 53.43 m

I hope this helps! If you still have questions let me know.

Bryce