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(D) Let the height of the pole be $h$. The ball is projected at an angle $	heta = 45^{\circ}$.Let $\alpha$ be the angle of elevation of the top of th

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$\frac{d_1 d_2}{d_1+d_2}$

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(D) Let the height of the pole be $h$. The ball is projected at an angle $	heta = 45^{\circ}$.Let $\alpha$ be the angle of elevation of the top of the pole from the point of projection,and $\beta$ be the angle of depression of the point where the ball lands from the top of the pole.From the geometry of the trajectory,we have $	an \alpha = \frac{h}{d_1}$ and $	an \beta = \frac{h}{d_2}$.For a projectile passing through a point $(x, y)$,the equation of trajectory is $y = x 	an 	heta \left(1 - \frac{x}{R}ight)$,where $R$ is the range.Alternatively,using the property of projectile motion,$	an 	heta = 	an \alpha + 	an \beta$.Substituting the values: $	an 45^{\circ} = \frac{h}{d_1} + \frac{h}{d_2}$.Since $	an 45^{\circ} = 1$,we have $1 = h \left(\frac{1}{d_1} + \frac{1}{d_2}ight) = h \left(\frac{d_1 + d_2}{d_1 d_2}ight)$.Therefore,the height of the pole is $h = \frac{d_1 d_2}{d_1 + d_2}$.

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal from a distance $d_1$ from the foot of a pole. It just touches the top of the pole and then falls on the ground at a distance $d_2$ from the pole on the other side. The height of the pole is ...........

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