Answer :
Given that,
Horizontal velocity of the object, v = 20 m/s
Height of the cliff, h = 125 m
We need to find the time that it takes the object to fall to the ground from the cliff is most nearly. It can be calculated using second equation of motion. Let us consider that the initial speed of the object is 0. So,
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
Here, a = g and u = 0
[tex]h=\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 125}{10}} \\\\t=5\ s[/tex]
So, the object will take 5 seconds to fall to the ground from the cliff.
The time taken for the object to fall to the ground is 5 s
From the question given above, the following data were obtained:
- Horizontal velocity (u) = 20 m/s
- Height (h) = 125 m
- Acceleration due to gravity (g) = 10 m/s²
- Time (t) =?
The time taken for the object to fall to the ground can be obtained as follow:
[tex]t = \sqrt{ \frac{2h}{g}} \\ \\ t = \sqrt{ \frac{2 \times 125}{10}} \\ \\ t = 5 \: s \\ \\ [/tex]
Therefore, the time taken for the object to get to the ground is 5 s
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