A ball is projected at an angle of $45^{\circ}$ 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 distance $d_1+d_2$ from the point of projection, then ' $h$ ' is

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Aicharya · Accepted Answer

$R=d_1+d_2=\frac{u^2 \sin 90^{\circ}}{g}$ $\therefore \frac{\mathrm{u}^2}{\mathrm{~g}}=\mathrm{d}_1+\mathrm{d}_2$

Using, $y=x \tan \theta-\frac{\mathrm{gx}^2}{24^2 \cos ^2 \theta}$ $\therefore \mathrm{h}=\mathrm{d}_1 \tan 45^{\circ}-\frac{\mathrm{d}_1^2}{2\left(\mathrm{~d}_1+\mathrm{d}_2\right) \times \frac{1}{2}}=\frac{\mathrm{d}_1 \mathrm{~d}_2}{\mathrm{~d}_1+\mathrm{d}_2}$

A ball is projected at an angle of $45^{\circ}$ with the horizontal. It passes through a wall of height '…

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