KCET · Physics · Motion In Two Dimensions
The maximum range of a gun on horizontal plane is \(16 \mathrm{~km}\). If \(g=10 \mathrm{~ms}^{-2}\), then muzzle velocity of a shell is
- A \(160 \mathrm{~ms}^{-1}\)
- B \(200 \sqrt{2} \mathrm{~ms}^{-1}\)
- C \(400 \mathrm{~ms}^{-1}\)
- D \(800 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(400 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(R=16 \mathrm{~km}, g=10 \mathrm{~ms}^{-2}\)
As we know that,
maximum range, \(R_{\max }=\frac{u^{2}}{g}\)
\(\Rightarrow \quad 16 \times 10^{3}=\frac{u^{2}}{10}\)
\(\Rightarrow \quad u=\sqrt{16 \times 10^{3} \times 10}\)
\(=4 \times 10^{2}=400 \mathrm{~ms}^{-1}\)
As we know that,
maximum range, \(R_{\max }=\frac{u^{2}}{g}\)
\(\Rightarrow \quad 16 \times 10^{3}=\frac{u^{2}}{10}\)
\(\Rightarrow \quad u=\sqrt{16 \times 10^{3} \times 10}\)
\(=4 \times 10^{2}=400 \mathrm{~ms}^{-1}\)
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