At a height 0.4, m from the ground, the veloc | vedclass

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Question

$\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gh}$

or $\quad \mathrm{u}^{2}=\mathrm{v}^{2}+2 \mathrm{gh}$

or $\quad \mathrm{u}_{\mathrm{x}}^{2}+\math

Vedclass · Accepted Answer

$30$

Vedclass · Answer

$\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gh}$

or $\quad \mathrm{u}^{2}=\mathrm{v}^{2}+2 \mathrm{gh}$

or $\quad \mathrm{u}_{\mathrm{x}}^{2}+\mathrm{u}_{\mathrm{y}}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+2 \mathrm{gh}$

As $\quad v_{x}=u_{x}$

$	herefore \quad \mathrm{u}_{y}^{2}=\mathrm{v}_{y}^{2}+2 \mathrm{gh}$

or $\quad \mathrm{u}_{\mathrm{y}}^{2}=(2)^{2}+2 	imes 10 	imes 0.4=12$

or $\quad \mathrm{u}_{y}=\sqrt{12}=2 \sqrt{3} \mathrm{m} / \mathrm{s}$

and $\quad \mathrm{u}_{\mathrm{x}}=\mathrm{v}_{\mathrm{x}}=6 \mathrm{m} / \mathrm{s}$

$	an 	heta=\frac{\mathrm{u}_{\mathrm{y}}}{\mathrm{u}_{\mathrm{x}}}=\frac{2 \sqrt{3}}{6}=\frac{1}{\sqrt{3}}$

$	herefore \quad 	heta=30^{\circ}$

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