JEE Mains · Physics · STD 11 - 3.2 motion in plane
The trajectory of a projectile near the surface of the earth is given as\( y = 2x -9x^2\). If it were launched at an angle \(\theta_0\) with speed \(v_0\) then \((g = 10\, ms^{-2}\))
- A \({\theta _0} = {\cos ^{ - 1}}\,\left( {\frac{1}{{\sqrt 5 }}} \right)\) and \({v_0} = \frac{5}{3}\,m{s^{ - 1}}\)
- B \({\theta _0} = {\cos ^{ - 1}}\,\left( {\frac{2}{{\sqrt 5 }}} \right)\) and \({v_0} = \frac{3}{5}\,m{s^{ - 1}}\)
- C \({\theta _0} = {\sin ^{ - 1}}\,\left( {\frac{2}{{\sqrt 5 }}} \right)\) and \({v_0} = \frac{3}{5}\,m{s^{ - 1}}\)
- D \({\theta _0} = {\sin ^{ - 1}}\,\left( {\frac{1}{{\sqrt 5 }}} \right)\) and \({v_0} = \frac{5}{3}\,m{s^{ - 1}}\)
Answer & Solution
Correct Answer
(A) \({\theta _0} = {\cos ^{ - 1}}\,\left( {\frac{1}{{\sqrt 5 }}} \right)\) and \({v_0} = \frac{5}{3}\,m{s^{ - 1}}\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} Equation\,of\,trajectory\,is\,given\,as\\ \,\,\,\,\,\,\,\,\,\,y = 2x - 9{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( A \right)\\ \,\,\,Comparing\,with\,equation\,\,\,\,\,\,\,\,\,\,\,\,\\ \,\,\,\,\,\,\,\,\,y = \,X\,\tan \,\theta -…
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