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  • A projectile is given an initial velociy of , where is along the ground and is along the vertical. If g = 10m/s2 , the equation of its tr

\\\text{A projectile is given an initial velocity of } \hat{\mathrm{i}}+2 \hat{\mathrm{j}} \ \mathrm{m} / \mathrm{s} where \hat{\mathrm{i}} is along the ground and \hat{\mathbf{j}} is along the vertical. If \mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2}, the equation of its trajectory is

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\begin{array}{l} \mathrm{U}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{m} \\ \mathrm{U}_{x}=\mathrm{im} \\ \mathrm{x}=\mathrm{S}_{\mathrm{x}}=\mathrm{U}_{x} \cdot \mathrm{t} \\ \mathrm{x}=\mathrm{S}_{\mathrm{x}}=1 \mathrm{t}=\mathrm{t} \ldots \ldots \ldots \ldots \text { (i) } \end{array}

\\\mathrm{U}_{y}=2 \hat{\mathrm{j}} \\ \mathrm{y}=\mathrm{S}_{y}-\mathrm{Uyt}+\frac{1}{2} \mathrm{at}^{2} \\ \mathrm{S}_{y}=2 \mathrm{t}-\frac{1}{2} \times 10 \mathrm{t}^{2} \\ \mathrm{y}=\mathrm{S}_{\mathrm{y}}=2 \mathrm{t}-5 \mathrm{t}^{2} \ldots \ldots \ldots \text { (ii) }

\begin{aligned} &\text {From equation (i) }\\ &y=2(x)-5 x^{2}\\ &5 x^{2}-2 x+y=0 \end{aligned}

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