Question

A ball rolls off the edge of a horizontal table top 4 m high. If it strikes the floor at a point 5 m horizontally away from the edge of the table, what was its speed at the instant it left the table?

Answer 1

We should know the three equations of motions.
We should know in horizontal direction all the quantity of y direction will be zero and for vertical motion all the quantities having horizontal components should be zero.

Complete step by step answer:
Motion is the state of change in position of an object over time. It is described in terms of displacement, distance, velocity, acceleration, time and speed.

The relations between these quantities are known as the equations of motion.In case of uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the components like displacement $\left( s \right),$velocity $\left( {initial{\text{ }}and{\text{ }}final} \right),$
time $\left( t \right)$ and acceleration $\left( a \right).$Therefore they can only be applied when acceleration is constant and motion is a straight line. The three equations are,

Where, $s{\text{ }} =$ displacement; $u{\text{ }} =$ initial velocity; $v{\text{ }} =$ final velocity; $a{\text{ }} =$ acceleration; $t{\text{ }} =$ time of motion. These equations are referred as ${S_y} = {U_y} + \dfrac{1}{2}{a_y}{t^2}$ equations where $SUVAT$ stands for displacement $\left( s \right),$initial velocity $\left( u \right),$ final velocity $\left( v \right),$ acceleration $\left( a \right).$and time $\left( T \right)$

We know that, the displacement equation of equations of motion is,

${S_y} = {U_y} + \dfrac{1}{2}{a_y}{t^2}$
${S_y} = 0 \times 1 + \dfrac{1}{2}\left( { - 9.81} \right){t^2}$
$t = \sqrt {\dfrac{{2 \times 1}}{{9.81}}}$
$t = 0.95wc$
Horizontal direction,
${S_y} = 5m$ ${U_Y} = v$ ${a_{y = 0}}$
$\implies {S_y} = {U_Y}t + \dfrac{1}{2}{a_x}{t^2}$
$\implies S = U\left( {0.9} \right) + \dfrac{1}{2}\left( 0 \right) \times t$
$\implies U = \dfrac{5}{{0.9}}$
$\therefore U = 5.55\dfrac{M}{S}$

Note:
We should take care of numerical errors.
We should convert all quantities into its SI unit system.