A ball is thrown vertically upward from the ground with a speed of 25-

A ball is thrown vertically upward from the ground with a speed of 25-2 m/s. The ball will reach the highest point of its journey in

[amp_mcq option1=”5·14 s” option2=”3·57 s” option3=”2·57 s” option4=”1·29 s” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2016

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Using the kinematic equation $v = u + at$, where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $t$ is time.

Given initial velocity $u = 25.2 \text{ m/s}$ upwards, and acceleration due to gravity $a = -9.8 \text{ m/s}^2$ (assuming upward is positive). At the highest point, the velocity $v = 0 \text{ m/s}$.

Substituting the values into the equation: $0 = 25.2 + (-9.8)t$.
Solving for $t$: $9.8t = 25.2$, so $t = 25.2 / 9.8 = 252 / 98 = 126 / 49 = 18 / 7 \approx 2.57$ seconds.

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