Question

A small ball rolls horizontally off the edge of a tabletop that is 1.23 m high. It strikes the floor at a point 1.49 m horizontally away from the edge of the table. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves ?

Answer 1

$\sf{\underline{We\:know\:that:}}$

The ball starts with the vertical velocity.

$\sf{\underline{That\:is:}}$

$\boxed{\sf{v_{0y} = 0} }$ falls freely to the ground, from a distance of 1.23 m high.

$\sf{\underline{So:}}$

The ball falls from the position of:

$\boxed{\sf{y_{0} = 1.23 \: m}}$

$\sf{\underline{We\:know\:that:}}$

$\boxed{\sf{y = y_{0} + v_{0y}t - \frac{1}{2} {gt}^{2}}}$

$\sf{\underline{That\:is:}}$

$\boxed{\sf{y = - \frac{1}{2} {gt}^{2}}}$

$\sf{\underline{Here:}}$ y = - 1.23 m

$\sf{\underline{Therefore:}}$

$\sf{\underline{The\:time\:taken\:to\:fall:}}$

$\implies$ $\sf{t ^{2} = \frac{2(1.23)}{9.8} }$

$\implies$ $\sf{t ^{2} = \frac{2.46}{9.8}}$

$\implies$ $\sf{{t}^{2} = 0.25}$

$\implies$ $\sf{t = \sqrt{0.25} }$

$\implies$ $\sf{t = 0.5}$

$\sf{\underline{So:}}$ $\boxed{\sf{t = 0.5\:s}}$

$\sf{\underline{Now:}}$

We can easily calculate the initial horizontal velocity of the ball.

$\sf{\underline{We\:know\:that}}$

$\boxed{\sf{x = x_{0} + v_{0x} t}}$

$\sf{\underline{So:}}$

$\implies$ $\sf{v_{0x} = \frac{x}{t}}$

$\implies$ $\sf{v_{0x} = \frac{1.49}{0.5}}$

$\implies$ $\sf{v_{0x} = 2.98 \: m/s}$

$\sf{\underline{Final\:answer:}}$

(a) $\sf{0.5\:s}$

(b) $\sf{2.98 \: m/s}$

Answer 2

Using h= ½ gt², we have

$h_{AB}= \frac{1}{2} gt^2_{AC}$

$or \: t_{AC} = \sqrt\frac{2h_{AB}}{g}$

$= \sqrt \frac{2 \times 4}{9.8} = 0.9s$

$Further, BC=vt_{AB}$

$or \: v= \frac{BC}{t_{AC}} = \frac{5.0}{0.9} =5.55m/s$