A ball is projected at an angle 45^{circ} wit | vedclass

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(B) The equation of the trajectory of a projectile is given by:$y = x 	an 	heta \left(1 - \frac{x}{R}ight)$Where $R$ is the horizontal range of th

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$h = \frac{d_1d_2}{d_1 + d_2}$

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(B) The equation of the trajectory of a projectile is given by:$y = x 	an 	heta \left(1 - \frac{x}{R}ight)$Where $R$ is the horizontal range of the projectile.Given the total horizontal distance covered is $R = d_1 + d_2$ and the angle of projection is $	heta = 45^{\circ}$.At horizontal distance $x = d_1$,the height of the ball is $y = h$.Substituting these values into the trajectory equation:$h = d_1 	an 45^{\circ} \left(1 - \frac{d_1}{d_1 + d_2}ight)$Since $	an 45^{\circ} = 1$:$h = d_1 \left(\frac{d_1 + d_2 - d_1}{d_1 + d_2}ight)$$h = d_1 \left(\frac{d_2}{d_1 + d_2}ight)$$h = \frac{d_1 d_2}{d_1 + d_2}$

$A$ ball is projected at an angle $45^{\circ}$ with the horizontal. It passes through a wall of height $h$ at a horizontal distance $d_1$ from the point of projection and strikes the ground at a horizontal distance $(d_1 + d_2)$ from the point of projection. Then $h$ is:

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