Question

The time of ascent is equal to

• A. Time of journey
• B. Time of descent
• C. Both
• D. None of these

Answer 1

The correct option is B

Time of descent

Step 1: Assuming the terms

The route traveled by a body hurled vertically upward is equal to the path taken by the body during descent. Furthermore, whenever air resistance is low, the velocity at the beginning is equivalent to the velocity at the end.

For ascent,

Time taken is ${\mathrm{t}}_1$.

Initial velocity is u.

The final velocity is $\mathrm{v}=0{\mathrm{ms}}^{-1}$.

Acceleration due to gravity is $\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$.

For descent,

Time taken is ${\mathrm{t}}_2$.

Initial velocity is $\mathrm{u}=0{\mathrm{ms}}^{-1}$.

The final velocity is v.

Acceleration due to gravity is $\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$.

Step 2: Finding the time of ascent

By using the first equation of motion $v=u+at$, where $a=-g$

we got the formula

${\mathrm{t}}_1=\frac{\mathrm{v}-\mathrm{u}}{-\mathrm{g}}$

By substituting the values,

$$\begin{gathered} =\frac{0-\mathrm{u}}{\mathrm{g}} \\ =\frac{\mathrm{u}}{\mathrm{g}} \end{gathered}$$

Step 3: Finding the time of descent

${\mathrm{t}}_2=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{g}}$

By substituting the values,

$$\begin{gathered} =\frac{\mathrm{v}-0}{\mathrm{g}} \\ =\frac{\mathrm{v}}{\mathrm{g}} \end{gathered}$$

Here we can say that,

$\mathrm{The}\mathrm{velocity}\mathrm{with}\mathrm{which}\mathrm{the}\mathrm{body}\mathrm{is}\mathrm{thrown}=\mathrm{The}\mathrm{velocity}\mathrm{with}\mathrm{which}\mathrm{the}\mathrm{body}\mathrm{hits}\mathrm{the}\mathrm{ground}.$

or,

${\mathrm{t}}_1={\mathrm{t}}_2$

Therefore, option B is the correct choice.