TS EAMCET · Physics · Motion In Two Dimensions
A particle is projected from the ground with some initial velocity making an angle of \(45^{\circ}\) with the horizontal. If it reaches a height of \(7.5 \mathrm{~m}\) above the ground, while it travels a horizontal distance of \(10 \mathrm{~m}\) from the point of projection, then the initial speed of particle is (assume, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(10 \mathrm{~m} / \mathrm{s}\)
- B \(20 \mathrm{~m} / \mathrm{s}\)
- C \(15 \mathrm{~m} / \mathrm{s}\)
- D \(25 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(20 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
The given situation is as shown in figure, Let the particle reaches point \(P\) in time \(t\). Then, horizontal distance, \(x=u \cos \theta t \Rightarrow 10=u \cos 45^{\circ} \times t\) \(\Rightarrow \quad t=\frac{10 \sqrt{2}}{u}\) and vertical distance,…
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