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Answer:
[tex]6.42\ \text{m/s}[/tex]
Explanation:
v = Final velocity
u = Initial velocity = 0
s = Displacement of Tarzan's feet = 2.1 m
a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the kinematic equations we get
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 2.1+0}\\\Rightarrow v=6.42\ \text{m/s}[/tex]
At the instant just before Tarzan's feet touch the ground the velocity of his feet are [tex]6.42\ \text{m/s}[/tex].
The speed at an instant just before Tarzen's feet touch the ground will be
[tex]V=6.42\ \dfrac{m}{s}[/tex]
The speed of any body is defined as the movement of the body with respect to the time.
It is given in the question that
Mass of the Tarzen m=100 kg
The distance of the Tarzen's feet from the ground s=2.1 m
The acceleration due to gravity [tex]a=9.81\ \frac{m}{s^2}[/tex]
The speed of the Tarzen will be calculated by applying the equation of motion.
[tex]v^2=u^2+2as[/tex]
Since initial velocity is zero u=0
[tex]v^2=2as[/tex]
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2\times 9.81\times 2.1} =6.42 \ \frac{m}{s}[/tex]
Thus the speed at an instant just before Tarzen's feet touch the ground will be
[tex]V=6.42\ \dfrac{m}{s}[/tex]
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