Question

When a ball is thrown vertically upward from the ground, it crosses a point which is at a height of $25\mathrm{m}$ twice in an interval of $4\mathrm{seconds}$. At what velocity was the ball thrown?

Answer 1

Step 1: Given data

Height $\left(\mathrm{h}\right)=$ $25\mathrm{m}$

Time $\left(\mathrm{t}\right)=$ $4$ seconds.

Step 2: Formula used

The velocity ${\mathrm{V}}^2={\mathrm{u}}^2+2\mathrm{gh}$

$\mathrm{V}=$Final velocity

$\mathrm{u}=$Initial velocity

$\mathrm{g}=$Gravitational force

$\mathrm{h}=$Height

Step 3: Determine the time

Consider the `$\mathrm{A}$` factor on the ground. Form factor, one ball is thrown vertically upwards to factor `$\mathrm{B}$`. From factor `$\mathrm{A}$` to `$\mathrm{B}$` top is $25\mathrm{m}$ after achieving at factor `$\mathrm{A}$` ball crosses this factor two times at a c program language period of $4\sec$.

It approaches ball blanketed factor `$\mathrm{B}$` to `$\mathrm{C}$` at the time $2\sec$ after which go back returned at identical factor with time$2\sec$.

Let us provide an explanation with the assistance of the subsequent diagram :

Let,

Velocity at point s ${\mathrm{B}}^{\text{'}}$is$\mathrm{v}$.

Velocity at the point ${\mathrm{C}}^{\text{'}}$ is ${\mathrm{v}}_1$.
The height from point ' $\mathrm{B}$ ' to$\mathrm{C}$
We know that,
${\mathrm{h}}_{\mathrm{t}}=\frac{1}{2}{\mathrm{gt}}^2$ (Velocity at the point $\mathrm{C}$ is $0$ )
$\mathrm{g}=9.8\mathrm{m}/\mathrm{s}$ (i.e. nearly equal to $10\mathrm{m}/\mathrm{s}$ )
$\mathrm{t}=2\sec$

Step 4: Determine the velocity of the ball thrown

Put the values of $\mathrm{g}$ and we get the value of
$\mathrm{h}=\frac{1}{2}\times 10\times 2^2$
$\mathrm{h}=20\mathrm{m}$

Total height from point ' ${\mathrm{A}}^{\text{'}}$ to $\mathrm{C}$ is
${\mathrm{h}}_{\max }={\mathrm{h}}_{\mathrm{AB}}+{\mathrm{h}}_{\mathrm{BC}}$

${\mathrm{h}}_{\max }=25+20$

${\mathrm{h}}_{\max }=45\mathrm{m}$

Velocity is found from:
${\mathrm{V}}^2={\mathrm{u}}^2+2\mathrm{gh}$

Here,

$\mathrm{V}$ is the velocity at point ' $\mathrm{C}$ '.

$\mathrm{U}$ is the velocity at point ' $\mathrm{A}\text{'},\mathrm{g}$ is gravity.

h is the height from point ' $\mathrm{A}$ ' to '$\mathrm{C}$'.

Velocity at point ' ${\mathrm{C}}^{\text{'}}$ is zero
${\mathrm{u}}^2=2\mathrm{gh}$

${\mathrm{u}}^2=2\times 10\times 45$

$\mathrm{u}=\sqrt{900}$

$\mathrm{u}=30\mathrm{m}/\mathrm{s}$

A ball is thrown vertically upward from the ground, it crosses a point that is at a height of $25\mathrm{m}$ twice in an interval of 4 seconds at the velocity of $30\mathrm{m}/\mathrm{s}$.