Respuesta :
Answer:
t = 0.97 s
Explanation:
Given that,
height, h = 4.7 m
We need to find the time taken by the object to fall 4.7 m.
Let it is t. Using second equation of kinematics to solve it such that,
[tex]h=ut+\dfrac{1}{2}gt^2[/tex]
u is initial velocity, u = 0
So,
[tex]h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}}\\\\t=\sqrt{\dfrac{2\times 4.7}{9.8}}\\\\t=0.97\ s[/tex]
So, the required time is equal to 0.97 seconds.
The time it would take the object is 0.98 secs
From the question,
We are to determine how much time it would take for an object to fall 4.7 meters. That is, the time it would take the object to fall from a height 4.7 meters.
From one of the equations of motions for free falling object.
[tex]h = ut + \frac{1}{2} gt^{2}[/tex]
Where h is the height
u is the initial velocity
t is the time
and g is the acceleration due to gravity (Take g = 9.8 m/s²)
Since, the object is dropped from rest, the initial velocity, u, is 0 m/s
From the given information,
h = 4.7 m
Putting the parameters into the equation, we get
[tex]4.7 = 0(t) +\frac{1}{2} (9.8)t^{2}[/tex]
[tex]4.7 = 0 + 4.9t^{2}[/tex]
[tex]4.7 = 4.9t^{2}[/tex]
∴ [tex]t^{2} = \frac{4.7}{4.9}[/tex]
[tex]t^{2}= 0.9592[/tex]
[tex]t = \sqrt{0.9592}[/tex]
t = 0.98 secs
Hence, the time it would take the object is 0.98 secs
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