A 0.5 kg ball is thrown up into the air with a velocity of 8 m/s. What is the maximum height this ball reaches?

Since, the ball is projected vertically upward with the initial velocity of 8 m/s. So, through this information, we can conclude that when the ball will reach the maximum height it's final velocity will be zero.

We know that, the direction of motion of the ball is in the opposite direction of acceleration due to gravity. So, the acceleration due to gravity will be negative, that is -10 m/s².

So, by using a required formula, we'll find the maximum height attained by the ball.

[tex] \qquad \qquad \qquad {\boxed{\sf{\blacksquare\ {\red{2as\ =\ v^2\ -\ u^2}}}}} [/tex]

By using the above formula, which is the third equation of motion, we'll find out the maximum height.

[tex] \red {\frak{Where}} \begin{cases} &\sf {a\ denotes\ acceleration.} \\ &\sf{s\ denotes\ distance\ travelled\ (Height).} \\ &\sf{v\ denotes\ final\ velocity.} \\ &\sf{u\ denotes\ initial\ velocity.} \end{cases}[/tex]

So, by using the required information provided above, we'll calculate the maximum height attained by the ball.

[tex] \bigstar {\underline{\underline{\red{\sf{Finding\ the\ maximum\ height:-}}}}} [/tex]

[tex] \sf \dashrightarrow {2as\ =\ v^2\ -\ u^2} \\ \\ \\ \sf \dashrightarrow {2 \times (-10\ m/s^2) \times max^{m}_{h}\ =\ (0\ m/s)^2\ -\ (8\ m/s)^2} \\ \\ \\ \sf \dashrightarrow {-20\ m/s^2 \times max^{m}_{h}\ =\ -64\ m^2/s^2} \\ \\ \\ \sf \dashrightarrow {max^{m}_{h}\ =\ \dfrac{\cancel{-64\ m/s^2}}{\cancel{-20\ m/s^2}}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{max^{m}_{h}\ =\ 3.2\ m}}}}_{\sf \blue {\tiny{Maximum\ Height}}}} [/tex]

∴ Hence, the required maximum height attained by the ball is 3.2 m.

[tex]\rule{200}{3}[/tex]