Question

A ball rolls off a level table with a speed of 2.2 m/s. The table is 1.2 meters high. How far away from the base of the table does the ball land?

Answer 1

The base of table is "1.08856 m" far away from the ball land.

Given:

Distance travelled by ball,

• S = 1.2 m

Initial velocity,

• u = 0 m/s

Acceleration,

• a = 9.8 m/s²

Constant speed,

• 2.2 m/s

As we know the formula,

→ $S = ut+\frac{1}{2}at^2$

By substituting the values, we get

→ $1.2=\frac{1}{2}\times 9.8t^2$

→ $1.2 = 4.9 t^2$

→    $t = \sqrt{\frac{1.2}{4.9} }$

→      $= 0.4948 \ seconds$

Let,

• Horizontal distance will be "x".
• Time = t

→ $x = 2.2\times t$

     $= 2.2\times 0.4948$

     $= 1.08856 \ m$

Thus the solution above is right.

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