Answer :
Answer:
20 meters
Explanation:
Kinematics
Kinematics is the study of basic motion that includes,
- initial and final velocity (in meters per second) [tex]v_0, \:v_f[/tex]
- acceleration (in meters per second squared) [tex]a[/tex]
- displacement (in meters) [tex]\rm \Delta x\: or\: \Delta y[/tex]
- time (in seconds) [tex]t[/tex].
When solving a kinematics problem, not all of these variables/features needs to be defined to find the value of asked variable but, three must be found.
To find the value the problem asks for, examine and choose one kinematics equation to utilize.
Depending on the variables given by the problem and what the problem is asking for, some rearranging of equation may be needed. See the attached image to view the three basic kinematics equations.
Applying Kinematics
The problem says that the object is thrown and is caught after 2 seconds, meaning that time or t = 2s.
Since the object is thrown in the y-axis (up and down), it has gravity due to earth or [tex]\pm10\frac{m}{s^2}[/tex], since the acceleration is a vector, the direction or the sign can be chosen by the individual's choice. So, let acceleration or a = 10 [tex]\frac{m}{s^2}[/tex].
To find the last bit of information, we must use intuition. Before the object is thrown, the object must be at rest, meaning that its initial velocity is zero or, [tex]v_0=0\frac{m}{s}[/tex].
The problem asks for the height that the object rose, or the displacement in the y-direction, [tex]\Delta y[/tex].
Inspecting the three kinematics equations and the variables that we have, we can see that second equation can be utilized.
Replacing x and [tex]x_0[/tex] with its y counterparts and rearranging the equation so all the y's are on one side,--which the difference between the y terms is [tex]\Delta y[/tex]--we have,
[tex]\Delta y = v_0t+\frac{1}{2}at^2[/tex].
Plugging in all the appropriate values and variables,
[tex]\Delta y= (0)(2)+\frac{1}{2} (10)(2)^2\\= 5(2)^2\\=5(4)\\\Longrightarrow\Delta y=20[/tex].
The object rose 19.6m above its initial height.
The object rose above its initial height, we can use the kinematic equation for vertical motion under constant acceleration:
h = [tex]v_i t + \frac{1}{2} a t^2[/tex]
where:
- - h is the displacement (height above initial position),
- - [tex]\( v_i \)[/tex] is the initial velocity (when the object was thrown upward),
- - a is the acceleration due to gravity (assuming downward direction),
- - t is the time.
Given:
- The object is thrown straight upwards and caught again at the same height, meaning its displacement is zero,
- The time taken for the entire motion (up and down) is 2 seconds.
Since the displacement is zero, we have:
h = 0
Thus, we're left with the equation for the entire vertical motion:
[tex]\[ v_i t + \frac{1}{2} a t^2[/tex] = 0
The time taken for the entire motion is 2 seconds, so t = 2 seconds.
- [tex]\[ v_i \times 2 + \frac{1}{2} a \times (2)^2[/tex] = 0
- 2[tex]v_i[/tex] + 2a = 0
- [tex]v_i[/tex] = -a
The negative sign indicates that the initial velocity is in the opposite direction of the acceleration due to gravity. Now, we need to find the initial velocity [tex]\( v_i \)[/tex] in terms of the acceleration due to gravity a.
When the object is thrown upwards, its velocity decreases until it reaches its highest point, where its velocity becomes zero before it starts to fall back down. Therefore, the initial velocity [tex]\( v_i \)[/tex] is equal in magnitude but opposite in direction to the final velocity when the object is at its highest point.
At the highest point, the object's velocity is zero. So, we can use the equation of motion:
[tex]v_f[/tex] = [tex]v_i[/tex] + at
where:
- - [tex]\( v_f \)[/tex] is the final velocity (zero at the highest point),
- - [tex]\( v_i \)[/tex] is the initial velocity,
- - a is the acceleration due to gravity (negative, as it acts downwards),
- - t is the time taken to reach the highest point.
Plugging in the values:
0 = [tex]v_i[/tex] + [tex](-9.8 \, \text{m/s}^2) \times 2[/tex]
[tex]v_i[/tex] = 19.6 m/s
The maximum height reached using the equation:
- h = [tex]\frac{v_i^2}{2a}[/tex]
- h = [tex]\frac{(19.6 \, \text{m/s})^2}{2 \times (-9.8 \, \text{m/s}^2)}[/tex]
- h = [tex]\frac{384.16 \, \text{m}^2/\text{s}^2}{-19.6 \, \text{m/s}^2}[/tex]
- h = -19.6 m.
