The trajectory of projectile, projected from the ground is given by y=x−x220. Where x and y are measured in meter. The maximum height attained by the projectile will be

y=x−x220y=tan⁡θx−g2u2cos2⁡θx2tan⁡θ=1⇒θ=45∘g2u2cos2⁡θ=120gu2=120u2g=20Hmax=u2sin2⁡θ2g=u24gHmax=14(20)=5m

The trajectory of projectile, projected from the ground is given by y=x−x220. Where x and y are measured in meter. The maximum height attained by the projectile will be

y=x−x220y=tan⁡θx−g2u2cos2⁡θx2tan⁡θ=1⇒θ=45∘g2u2cos2⁡θ=120gu2=120u2g=20Hmax=u2sin2⁡θ2g=u24gHmax=14(20)=5m

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The trajectory of projectile, projected from the ground is given by $y = x - \frac{x^{2}}{20}$ . Where x and y are measured in meter. The maximum height attained by the projectile will be

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200 m

10 $\sqrt{2}$ m

5 m

10 m

answer is B.

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Detailed Solution

$\begin{array}{l} y = x - \frac{x^{2}}{20} \\ y = \tan θx - \frac{g}{2 u^{2} \cos^{2} θ} x^{2} \\ \tan θ = 1 \Rightarrow θ = 45^{\circ} \\ \frac{g}{2 u^{2} \cos^{2} θ} = \frac{1}{20} \\ \frac{g}{u^{2}} = \frac{1}{20} \\ \frac{u^{2}}{g} = 20 \\ H_{max} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{u^{2}}{4 g} \\ H_{\max} = \frac{1}{4} (20) = 5 m \end{array}$