A ball rolls with a speed of $2.0 \frac{m}{s}$ $2.0m/s$ across a level table that is $1.0 m$ $1.0m$ above the floor. Upon reaching the edge of the table, it follows a parabolic path to the floor. How far along the floor is the landing spot from the table?

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A ball rolls with a speed of $2.0 \frac{m}{s}$ $2.0m/s$ across a level table that is $1.0 m$ $1.0m$ above the floor. Upon reaching the edge of the table, it follows a parabolic path to the floor. How far along the floor is the landing spot from the table?

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Answer 1 · 2016-03-08T21:25:49

We need to consider vertical and horizontal components.

Explanation:

If we consider the vertical component, we can determine how long the ball is in free fall. We can use:
$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$
S=1m
And u, initial velocity in vertical direction, is 0m/s
And we can take acceleration as 9.81 $m s^{- 2}$ $ms^-2$
$1 = \frac{1}{2} \cdot 9.8 \cdot t^{2}$ $1=1/2*9.8*t^2$

So t= 0.45s

In the horizontal direction, there is no acceleration so we can simply use $v e l o c i t y = \frac{S}{t}$ $velocity= S/t$
Hence distance along the floor from the table is $2 \times 0.45 = 0.9 m$ $2times0.45=0.9m$

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A ball rolls with a speed of $2.0 \frac{m}{s}$ $2.0m/s$ across a level table that is $1.0 m$ $1.0m$ above the floor. Upon reaching the edge of the table, it follows a parabolic path to the floor. How far along the floor is the landing spot from the table?

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Explanation:

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A ball rolls with a speed of 2.0ms2.0m/s across a level table that is 1.0m1.0m above the floor. Upon reaching the edge of the table, it follows a parabolic path to the floor. How far along the floor is the landing spot from the table?

1 Answer

Explanation:

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A ball rolls with a speed of $2.0 \frac{m}{s}$ $2.0m/s$ across a level table that is $1.0 m$ $1.0m$ above the floor. Upon reaching the edge of the table, it follows a parabolic path to the floor. How far along the floor is the landing spot from the table?