Question

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

Answer 1

Hint:To solve this problem, we will first use the formula for free fall. By this, we will first find the time taken by the ball to touch the floor. After that by using the definition of speed, we will use this time and the horizontal distance covered by it to find our answer.

Formulas used:
$h = \dfrac{1}{2}g{t^2}$
where, $h$ is the height, $g$ is the acceleration due to gravity and $t$ is the time.
$v = \dfrac{d}{t}$
where, $v$ is the speed,$d$ is the distance covered and $t$ is the time.

Complete step by step answer:
Here, we are given that A ball rolls off the edge of the horizontal table top $4m$ high. If it strikes the floor at a point horizontally away from the edge of the table.We will take $h = 4m$, $g = 10m/{s^2}$ and $d = 5m$. Now, first we will apply the formula for free fall of the ball and find the time required by the ball to strike the floor.
$
h = \dfrac{1}{2}g{t^2} \\
\Rightarrow {t^2} = \dfrac{{2h}}{g} \\
\Rightarrow {t^2} = \dfrac{{2 \times 4}}{{10}} \\
\Rightarrow {t^2} = 0.8 \\
\Rightarrow t = 0.89 \approx 0.9\sec \\
 $
We will now find the speed at the instant the ball left the table.
$\therefore v = \dfrac{d}{t} = \dfrac{5}{{0.9}} = 5.55m/s$
Thus, the speed of the ball at the instant it left the table is $5.55m/s$.

Note:Here, we have used the concept of free fall of the ball. Free fall is defined as a situation when a body is moving only under the influence of the earth’s gravity. Since external force is acting on the ball, the motion will be accelerated. This free-fall acceleration is also known as acceleration due to gravity which we have used in this problem.