For the body that falls freely from rest to Earth, we have:
a. The displacement from t=0 to t=4 s is 78.5 m.
b. The time for it to reach a speed of 35 m/s is 3.57 s.
c. The time required for it to fall 400m is 9.03 s.
d. The body's speed after falling 140 m is 52.41 m/s.
a. The displacement is given by the vertical motion (it is free falling) and can be calculated with the following equation:
[tex] y_{f} = y_{i} + v_{i_{y}}t - \frac{1}{2}gt^{2} [/tex] (1)
Where:
[tex] y_{f}[/tex]: is the final height = 0
[tex] y_{i} [/tex]: is the initial height =?
[tex] v_{i_{y}}[/tex]: is the initial velocity in the y-direction = 0 (it falls from rest)
t: is the time = 4 s - 0 = 4 s
g: is the acceleration due to gravity = 9.81 m/s²
The displacement is:
[tex] y_{i} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(4 s)^{2} = 78.5 m [/tex]
Hence, the body's displacement is 78.5 m.
b. The time for it to reach a speed of 35 m/s is the following:
[tex] v_{f_{y}} = v_{i_{y}} + gt [/tex] (2)
Where:
[tex] v_{f_{y}} [/tex]: is the final speed = 35 m/s
The time is:
[tex] t = \frac{v_{f_{y}} - v_{i_{y}}}{g} = \frac{35 m/s}{9.81 m/s^{2}} = 3.57 s [/tex]
Then, it will reach a speed of 35 m/s in 3.57 s.
c. The time required for it to fall 400m can be founded with equation (1):
[tex] y_{f} = y_{i} - \frac{1}{2}gt^{2} [/tex]
[tex] 0 = 400 m - \frac{1}{2}*9.81 m/s^{2}*t^{2} [/tex]
[tex] t = \sqrt{\frac{2*400 m}{9.81 m/s^{2}}} = \sqrt{\frac{2*400 m}{9.81 m/s^{2}}} = 9.03 s [/tex]
Hence, the time required is 9.03 s.
d. The speed after falling 140 m can be calculated with the following equation:
[tex] v_{f}^{2} = v_{i}^{2} + 2g\Delta y [/tex]
[tex] v_{f} = \sqrt{2g\Delta y} = \sqrt{2*9.81 m/s^{2}*140 m} = 52.41 m/s [/tex]
Therefore, the speed after falling 140 m is 52.41 m/s.
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