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Sagot :
Given:
The mass of the ball, m=0.3 kg
The initial speed of the ball, u=32 m/s
The height, h=20 m
To find:
The speed of the ball when it is at a height of 20 m.
Explanation:
The ball will be moving under the influence of acceleration due to gravity which is directed downwards.
From the equation of motion,
[tex]v^2=u^2-2gh[/tex]Where v is the final velocity of the ball and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} v^2=32^2-2\times9.8\times20 \\ \implies v=\sqrt{32^2-2\times9.8\times20} \\ =25.1\text{ m/s} \end{gathered}[/tex]Final answer:
The velocity of the ball when it is 20 m above the releasing point is 25.1 m/s
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