AP EAMCET · PHYSICS · Motion In Two Dimensions
The equation for the trajectory of a projectile is \(y=\left(\frac{x}{\sqrt{3}}-\frac{x^2}{60}\right) m\). The velocity of projection of the projectile is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(8 \mathrm{~ms}^{-1}\)
- B \(40 \mathrm{~ms}^{-1}\)
- C \(16 \mathrm{~ms}^{-1}\)
- D \(20 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(20 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Coefficient of \(x^2\) in equation of trajectory \[ =\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta} \] So, \(\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\frac{1}{60} \Rightarrow \frac{10}{2 \mathrm{u}^2 \cos ^2 \theta}=\frac{1}{60}\)…
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