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)

$$\vec { v } _ { f y } = \vec { v } _ { 0 y } + \vec { a } _ { y } \delta t$$

$$\delta \vec { y } = \vec { v } _ { 0 y } \delta t + \frac { 1 } { 2 } \vec { a } _ { y } \delta t ^ { 2 }$$

$$\delta \vec { y } = ( \frac { \vec { v } _ { f y } + \vec { v } _ { 0 y } } { 2 } ) \delta t$$

Explanation:

Step1: Analyze the problem (free - fall motion)

Assume the object is in free - fall (starts from rest, so \(v_{0y} = 0\ m/s\), final velocity \(v_{fy}=50.0\ m/s\), acceleration \(a_y = g= 9.8\ m/s^{2}\) (acceleration due to gravity)). First, use the first equation \(v_{fy}=v_{0y}+a_y\Delta t\) to find the time of flight \(\Delta t\).
Given \(v_{0y} = 0\ m/s\), \(v_{fy}=50.0\ m/s\), \(a_y = 9.8\ m/s^{2}\)
From \(v_{fy}=v_{0y}+a_y\Delta t\), we can re - arrange for \(\Delta t\): \(\Delta t=\frac{v_{fy}-v_{0y}}{a_y}\)
Substitute the values: \(\Delta t=\frac{50.0 - 0}{9.8}=\frac{50.0}{9.8}\approx5.10\ s\)

Step2: Calculate displacement \(\Delta y\)

We can use the third equation \(\Delta y = (\frac{v_{fy}+v_{0y}}{2})\Delta t\)
Since \(v_{0y} = 0\ m/s\), \(v_{fy}=50.0\ m/s\), and \(\Delta t\approx5.10\ s\)
\(\Delta y=(\frac{50.0 + 0}{2})\times5.10\)
First, calculate \(\frac{50.0+0}{2}=25.0\ m/s\)
Then, \(\Delta y = 25.0\times5.10 = 127.5\ m\)
Or we can use the second equation \(\Delta y=v_{0y}\Delta t+\frac{1}{2}a_y\Delta t^{2}\)
Since \(v_{0y} = 0\), \(\Delta y=\frac{1}{2}\times9.8\times(5.10)^{2}=\frac{1}{2}\times9.8\times26.01 = 4.9\times26.01\approx127.45\ m\) (approximate to \(127\ m\) or \(128\ m\) depending on significant figures, but let's check the significant figures. The given \(v_{fy}=50.0\ m/s\) has 3 significant figures, \(g = 9.8\ m/s^{2}\) has 2 significant figures? Wait, no, if we take \(g = 9.81\ m/s^{2}\) for more accuracy. Let's recalculate with \(g = 9.81\ m/s^{2}\)
\(\Delta t=\frac{50.0}{9.81}\approx5.097\ s\)
Using \(\Delta y=\frac{v_{fy}+v_{0y}}{2}\Delta t=\frac{50.0 + 0}{2}\times5.097=25.0\times5.097 = 127.425\ m\approx127\ m\) (or if we consider the initial velocity is 0 and use \(v_{fy}^{2}=v_{0y}^{2}+2a_y\Delta y\) (a kinematic equation not given, but useful). From \(v_{fy}^{2}=v_{0y}^{2}+2a_y\Delta y\), since \(v_{0y} = 0\), \(\Delta y=\frac{v_{fy}^{2}}{2a_y}\)
\(\Delta y=\frac{(50.0)^{2}}{2\times9.81}=\frac{2500}{19.62}\approx127.4\ m\approx127\ m\) (or \(1.3\times 10^{2}\ m\) with two significant figures, but \(50.0\) has three, so maybe three significant figures: \(127\ m\) or \(128\ m\) depending on calculation)

Answer:

\(127\ m\) (or \(128\ m\) or \(1.3\times 10^{2}\ m\) depending on the value of \(g\) and significant figure considerations. If we use \(g = 9.8\ m/s^{2}\), \(\Delta y=\frac{50^{2}}{2\times9.8}=\frac{2500}{19.6}\approx127.5\ m\approx128\ m\))