The new tower musy be 4 times higher than the original tower
Explanation:
An object in free fall is an object acted upon gravity only, so it follows a uniformly accelerated motion, therefore its final velocity is given by the suvat equation:
[tex]v^2-u^2=2as[/tex]
where:
s is the vertical displacement of the ball (so, the height of the tower in this problem)
u is the initial vertical velocity (zero if the ball is dropped from rest)
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
v is the final velocity of the ball
We can rewrite the equation as
[tex]v^2 = 2as \rightarrow s=\frac{v^2}{2a}[/tex]
Since u = 0.
Therefore, we see that the height of the tower is proportional to the square of the final velocity of the ball.
In this problem, we want the ball to achieve a final speed which is twice the original speed, so
v' = 2v
Therefore, substituting into the equation,
[tex]s' = \frac{(2v)^2}{2g}=4(\frac{v^2}{2g})=4s[/tex]
this means that the new tower must be 4 times higher than the original tower.
Learn more about free fall:
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