Falling Objects (part 1)

A person standing on the edge of a high cliff throws a rock straight up with an initial velocity v₀ of 13 m/s.

(a) Calculate the position and velocity of the rock at 1.00, 2.00, and 3.00 seconds after it is thrown. Ignore the effect of air resistance.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Sketch the initial velocity and acceleration vectors and the axes.

Figure 3.12 Initial conditions for rock thrown straight up.

List the knowns: time t = 1.00 s, 2.00 s, and 3.00 s; initial velocity v₀ = 13 m/s; acceleration a = g = –9.80 m/s²; and position y₀ = 0 m

List the unknowns: y₁, y₂, and y₃; v₁, v₂, and v₃—where 1, 2, 3 refer to times 1.00 s, 2.00 s, and 3.00 s

Choose the equations.

$d = d_{0} + v_{0} t + \frac{1}{2} a t^{2} becomes y = y_{0} + v_{0} t - \frac{1}{2} g t^{2}$
3.11

$v = v_{0} + a t becomes v = v_{0} + - g t$
3.12

These equations describe the unknowns in terms of knowns only.
Solution
$\begin{matrix} y_{1} = 0 + (13.0 m/s) (1.00 s) + \frac{(- 9.80 {m/s}^{2}) {(1.00 s)}^{2}}{2} = 8.10 m \\ y_{2} = 0 + (13.0 m/s) (2.00 s) + \frac{(- 9.80 {m/s}^{2}) {(2.00 s)}^{2}}{2} = 6.40 m \\ y_{3} = 0 + (13.0 m/s) (3.00 s) + \frac{(- 9.80 {m/s}^{2}) {(3.00 s)}^{2}}{2} = - 5.10 m \\ v_{1} = 13.0 m/s + (- 9.80 {m/s}^{2}) (1.00 s) = 3.20 m/s \\ v_{2} = 13.0 m/s + (- 9.80 {m/s}^{2}) (2.00 s) = - 6.60 m/s \\ v_{3} = 13.0 m/s + (- 9.80 {m/s}^{2}) (3.00 s) = - 16.4 m/s \end{matrix}$

0 of 2 steps open

How did it go?