A ball is thrown vertically upwards. Prove that its time of ascent is equal to the time of descent.

Step 1: Given data

Final velocity during ascent $v=0$.

Initial velocity during descent $u=0$.

Acceleration during ascent $a=-g$.

Acceleration during descent $a=g$.

Step 2: Formula for calculating the time of journey

$$\begin{gathered} v=u+at \\ where, \\ v\to Finalvelocity \\ u\to Initialvelocity \\ a\to Acceleration \\ t\to Time \end{gathered}$$

Step 3: Calculating the time of ascent

$$\begin{gathered} v=u+at \\ \Rightarrow t=\frac{v-u}{a} \\ \Rightarrow t_{ascend}=\frac{0-u}{-g} \\ \Rightarrow t_{ascend}=\frac{u}{g} \end{gathered}$$

Step 4: Calculating the time of descent

$$\begin{gathered} v=u+at \\ \Rightarrow t=\frac{v-u}{a} \\ \Rightarrow t_{descent}=\frac{v-0}{g} \\ \Rightarrow t_{descent}=\frac{v}{g} \end{gathered}$$$v=u$

Step 5:Proving ${\mathit{t}}_{\mathbf{a}\mathbf{s}\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{=}\mathbf{}{\mathit{t}}_{\mathbf{d}\mathbf{e}\mathbf{s}\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{}}$
For a body that is thrown upwards, $v=u$.

Thus, $t_{\mathrm{ascent}}=t_{\mathrm{descent}}$.

Hence, for a ball that is thrown vertically upwards, the time of ascent is equal to the time of descent.