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(B) To find the initial velocity $(u)$ of the ball,we use the third equation of motion: $v^2 = u^2 + 2as$.At maximum height,the final velocity $(v)$ i

Vedclass · Accepted Answer

$44.2$

Vedclass · Answer

(B) To find the initial velocity $(u)$ of the ball,we use the third equation of motion: $v^2 = u^2 + 2as$.At maximum height,the final velocity $(v)$ is $0 \, m/s$.The acceleration due to gravity $(a)$ is $-g = -9.8 \, m/s^2$.The displacement $(s)$ is $100 \, m$.Substituting these values into the equation: $0^2 = u^2 + 2(-9.8)(100)$.$0 = u^2 - 1960$.$u^2 = 1960$.$u = \sqrt{1960} \approx 44.27 \, m/s$.Rounding to the nearest option,the speed is $44.2 \, m/s$.

$A$ ball is thrown up and attains a maximum height of $100 \, m$. It is thrown upwards with a speed of: (in $, m/s$)

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