Question

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?

Answer 1

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• A 0.5kg ball is thrown up with a velocity of 8m/s .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find :- }}}}$

• The maximum height attained by the ball.

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

Given that , a ball is projected vertically upward with initial velocity of 8 m/s . Here when the ball will reach the maximum height then its final velocity will be zero. And the acceleration due to gravity will be -10 m/s² , since the direction of motion of the ball is in opposite direction of acceleration due to gravity . Now using the Third equation of Motion , we have ,

$\sf\dashrightarrow 2as = v^2 - u^2 \\\\\\\sf\dashrightarrow 2 * (-10m/s^2) * h_{(max)} = (0m/s)^2-(8m/s)^2 \\\\\\\sf\dashrightarrow -20 m/s^2 * h_{(max)} = -64m^2/s^2 \\\\\\\sf\dashrightarrow h_{(max)} = \dfrac{-64\ m/s^2}{-20 \ m/s^2} \\\\\\\sf\dashrightarrow \underset{\blue{\sf Required \ Height }}{\underbrace{\boxed{\pink{\frak{ Height_{(maximum)}= 3.2\ m }}}}}$

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❒ Know More :-

• The maximum height obtained can directly be found out using the Formula , u²/2g , where u is the velocity of projection and g is the acceleration due to the gravity.

• The time of fight equals to 2u/g , where the time of acsent is equal to the time of descent and each equals to u/g .

• Here in the question , mass of the ball was given but velocity doesn't depends on the mass .

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