Question

an object lands on the ground with a speed of 50.0 m/s. what was the displacement of the object? (be sure your answer has the appropriate number of significant figures) (include unit: m, s, m/s, m/s^2) (put a space in between number and unit, example: 10.00 m/s^2)
\\( \vec { v } _ { f y } = \vec { v } _ { 0 y } + \vec { a } _ { y } \delta t \\)
\\( \delta \vec { y } = \vec { v } _ { 0 y } \delta t + \frac { 1 } { 2 } \vec { a } _ { y } \delta t ^ { 2 } \\)

Explanation:

Step1: Analyze the problem (free - fall assumption)

Assume the object is in free - fall, so the initial vertical velocity \(\vec{v}_{0y} = 0\ m/s\), the acceleration \(a_y=-g=- 9.8\ m/s^{2}\) (taking downwards as positive, we can also consider \(a_y = g = 9.8\ m/s^{2}\) if we take the direction of motion as positive), and the final vertical velocity \(v_{fy}=50.0\ m/s\). First, we use the equation \(v_{fy}=v_{0y}+a_y\Delta t\) to find the time of fall.
Since \(v_{0y} = 0\), the equation becomes \(v_{fy}=a_y\Delta t\). Solving for \(\Delta t\), we get \(\Delta t=\frac{v_{fy}}{a_y}\). If we take \(a_y = 9.8\ m/s^{2}\) (downwards as positive), then \(\Delta t=\frac{50.0\ m/s}{9.8\ m/s^{2}}\approx5.10\ s\).

Step2: Calculate the displacement

Now we use the displacement equation \(\Delta\vec{y}=\vec{v}_{0y}\Delta t+\frac{1}{2}a_y\Delta t^{2}\). Since \(\vec{v}_{0y} = 0\), the equation simplifies to \(\Delta y=\frac{1}{2}a_y\Delta t^{2}\). We can also use another kinematic equation \(v_{fy}^{2}=v_{0y}^{2}+2a_y\Delta y\). Since \(v_{0y} = 0\), we have \(\Delta y=\frac{v_{fy}^{2}}{2a_y}\).
Substitute \(v_{fy} = 50.0\ m/s\) and \(a_y=9.8\ m/s^{2}\) into the equation: \(\Delta y=\frac{(50.0\ m/s)^{2}}{2\times9.8\ m/s^{2}}=\frac{2500\ m^{2}/s^{2}}{19.6\ m/s^{2}}\approx127.55\ m\). Rounding to three significant figures (because \(50.0\) has three significant figures), we get \(\Delta y\approx128\ m\).

Answer:

128 m