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If a ball is dropped off a tall building and accelerates at 9.8 m/s^2 until reaching the ground at a speed of 55 m/s, how long will it take for the ball to reach the ground?

Sagot :

VirαtKσhliidn

Answer:

5.61s

Explanation:

[tex]\pink{\frak{Given}}\begin{cases} \textsf{ A ball is dropped off a tall building and accelerates at 9.8 m/s$^2$ .}\\\textsf{It reaches the ground at a speed of 55 m/s .} \end{cases}[/tex]

And we need to find out the time taken by the ball to reach the ground .

  • The initial velocity of the ball will be 0 since it is dropped .

Here we can use the First equation of motion , namely ;

[tex]\sf \longrightarrow v = u + at [/tex]

where the symbols have their usual meaning . Now substituting the respective values , we have,

[tex]\sf \longrightarrow 55m/s = 0m/s + 9.8 m/s^2(t)\\ [/tex]

[tex]\sf \longrightarrow 55m/s = 9.8m/s^2(t)\\ [/tex]

[tex]\sf \longrightarrow t = \dfrac{55m/s}{9.8m/s^2}\\[/tex]

[tex]\sf \longrightarrow \boxed{\bf time = 5.61 s }[/tex]

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