0.10 1.11 falling objects  (Page 2/8)

One-dimensional motion involving gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So we start by considering straight up and down motion with no air resistance or friction. These assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude $g$ . We will also represent vertical displacement with the symbol $y$ and use $x$ for horizontal displacement.

Calculating position and velocity of a falling object: a rock thrown upward

A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s . The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.

Strategy

Draw a sketch.

We are asked to determine the position $y$ at various times. It is reasonable to take the initial position $y_0$ to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so $a$ is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.

Since we are asked for values of position and velocity at three times, we will refer to these as $y_1$ and $v_1$ ; $y_2$ and $v_2$ ; and $y_3$ and $v_3$ .

Solution for Position $y_1$

1. Identify the knowns. We know that $y_0=0$ ; $v_0=\text{13}\text{.}\text{0 m/s}$ ; $a=-g=-9\text{.}{\text{80 m/s}}^2$ ; and $t=1\text{.}\text{00 s}$ .

2. Identify the best equation to use. We will use $y=y_0+v_{0}t+\frac{1}{2}{\text{at}}^2$ because it includes only one unknown, $y$ (or $y_1$ , here), which is the value we want to find.

3. Plug in the known values and solve for $y_1$ .

Discussion

The rock is 8.10 m above its starting point at $t=1\text{.}\text{00}$ s, since $y_1>y_0$ . It could be moving up or down; the only way to tell is to calculate $v_1$ and find out if it is positive or negative.

Solution for Velocity $v_1$

1. Identify the knowns. We know that $y_0=0$ ; $v_0=\text{13}\text{.}\text{0 m/s}$ ; $a=-g=-9\text{.}{\text{80 m/s}}^2$ ; and $t=1\text{.}\text{00 s}$ . We also know from the solution above that $y_1=8\text{.}\text{10 m}$ .

2. Identify the best equation to use. The most straightforward is $v=v_0-\text{gt}$ (from $v=v_0+\text{at}$ , where $a=\text{gravitational acceleration}=-g$ ).

3. Plug in the knowns and solve.

Discussion

The positive value for $v_1$ means that the rock is still heading upward at $t=1\text{.}\text{00}\text{s}$ . However, it has slowed from its original 13.0 m/s, as expected.