AP EAMCET · PHYSICS · Motion In Two Dimensions
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
- A \(h=\frac{2 d_1 d_2}{d_1+d_2}\)
- B \(h=\frac{d_1 d_2}{d_1+d_2}\)
- C \(h=\frac{\sqrt{2} d_1 d_2}{d_1+d_2}\)
- D \(h=\frac{d_1 d_2}{2\left(d_1+d_2\right)}\)
Answer & Solution
Correct Answer
(B) \(h=\frac{d_1 d_2}{d_1+d_2}\)
Step-by-step Solution
Detailed explanation
\(R=d_1+d_2=\frac{u^2 \sin 90^{\circ}}{g}\) \(\therefore \frac{\mathrm{u}^2}{\mathrm{~g}}=\mathrm{d}_1+\mathrm{d}_2\) Using, \(y=x \tan \theta-\frac{\mathrm{gx}^2}{24^2 \cos ^2 \theta}\)…
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