You throw a $20 \; N$ rock vertically into the air from ground level. You observe that when it is $15 \; m$ above the ground, it is travelling at $25 \; m/s$ upward. Use the work-energy theorem to find (i) its speed as it left the ground and (ii) its maximum height.

Given,

Weight of the rock $(W) = 20\;N$

Height $(h) = 15\;m$

Final speed$(v) = 25\;m/s$

Initial speed$(u) = \;?$

Maximum height $(h_{max}) = \;?$

(i) According to the Work-Energy theorem,

$W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Here, the rock is thrown vertically upward,

so, $W = mgh$

From equation (i) and (ii), we get

$ mgh = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

$2gh = v^2 - u^2$

$u^2 = v^2 - 2gh$

     $ = (25)^2 - 2 * (-9.8) * 15$

⇒ $u = 30.3\;m/s$

(ii) At the maximum height, velocity v = 0,

$mgh_{max} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

$gh_{max} = -\frac{1}{2}u^2$

$-\;9.8\;*\;h_{max} = -$$\frac{1}{2}u^2$

⇒ $h_{max} = \;$$\frac{(30.3)^2}{2 * 9.8}$$ = 46.8\;m$

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