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(A) Let $h$ be the height at which the kinetic energy becomes half of the initial kinetic energy.Initial kinetic energy $K_0 = \frac{1}{2} m v_0^2$.At

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(A) Let $h$ be the height at which the kinetic energy becomes half of the initial kinetic energy.Initial kinetic energy $K_0 = \frac{1}{2} m v_0^2$.At height $h$,kinetic energy $K = \frac{K_0}{2} = \frac{1}{4} m v_0^2$.Using the work-energy theorem or conservation of energy,the loss in kinetic energy equals the gain in potential energy:$K_0 - K = mgh$$\frac{1}{2} m v_0^2 - \frac{1}{4} m v_0^2 = mgh$$\frac{1}{4} m v_0^2 = mgh$$h = \frac{v_0^2}{4g}$Given $v_0 = 7 \, m/s$ and $g = 9.8 \, m/s^2$:$h = \frac{7^2}{4 	imes 9.8} = \frac{49}{39.2} = 1.25 \, m$.

$A$ body is projected vertically upwards with an initial velocity of $7 \, m/s$. At what height will its kinetic energy become half of its initial value (in $, m$)?

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