A ball is thrown at an angle theta and anothe | vedclass

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(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first ball,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the

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$15\,m, 65\,m$

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(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first ball,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the second ball,$H_2 = \frac{u^2 \sin^2(90^\circ - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.Given $u = 40\,m/s$ and $g = 10\,m/s^2$,we have $H_1 = \frac{1600 \sin^2 	heta}{20} = 80 \sin^2 	heta$ and $H_2 = 80 \cos^2 	heta$.Given $H_2 - H_1 = 50\,m$,so $80(\cos^2 	heta - \sin^2 	heta) = 50$.Using $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$,we get $80 \cos(2	heta) = 50$,so $\cos(2	heta) = \frac{5}{8} = 0.625$.Since $\cos(2	heta) = 1 - 2\sin^2 	heta$,then $2\sin^2 	heta = 1 - 0.625 = 0.375$,so $\sin^2 	heta = 0.1875$.Thus,$H_1 = 80 	imes 0.1875 = 15\,m$.Then $H_2 = H_1 + 50 = 15 + 50 = 65\,m$.

$A$ ball is thrown at an angle $\theta$ and another ball is thrown at an angle $(90^\circ - \theta)$ with the horizontal from the same point with the same speed $40\,m/s$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights.

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