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JEE Mains · Physics · STD 11 - 3.2 motion in plane

A projectile is given an initial velocity of \((\hat i + 2\hat j)\ ms^{-1}\), where \(\hat i\) is along the ground and \(\hat j\) is along the vertical. If \(g = 10\ m/s^2\) , the equation of its trajectory is

  1. A \(y= x- 5x^2\)
  2. B \(y= 2x- 5x^2\)
  3. C \(4y= 2x- 5x^2\)
  4. D \(4y= 2x- 25x^2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(y= 2x- 5x^2\)

Step-by-step Solution

Detailed explanation

\(\begin{array}{l} \vec u = \hat i + 2\hat j = {u_x}\hat i + {u_y}\hat j \Rightarrow u\cos \theta = 1,\\ u\sin \theta = 2\\ y = x\tan \theta - \frac{1}{2}\frac{{g{x^2}}}{{u_x^2}}\\ \therefore \,y = 2x - \frac{1}{2}g{x^2} = 2x - 5{x^2} \end{array}\)
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