Matching Projectile Trajectory Parameters – Rankers Physics

Ground to Ground Projectile: Practice Problem & Solution

A particle is moving in vertical plane (x-y plane) such that its trajectory is given by the equation, \(y = x - \frac{x^2}{80}\), where \(x\) & \(y\) are in metre. For this particle match column-I with column-II and tick the correct option. **Column I**: A. Horizontal range (in m) B. Angle of projection with horizontal (degree) C. Maximum height gained by particle (m) D. Speed of projection (\(\text{m s}^{-1}\)) **Column II**: (P) 45 (Q) 80 (R) 20 (S) \(20\sqrt{2}\)
A(S), B(R), C(Q), D(P)
A(P), B(Q), C(S), D(R)
A(Q), B(P), C(S), D(R)
A(Q), B(P), C(R), D(S)

Solution Explained:

To solve this problem, we apply the core principles of Ground to Ground Projectile. Understanding the underlying formula is key to arriving at the correct answer below:

Comparing the trajectory to \(y = xtan\theta - \frac{gx^2}{2u^2cos^2\theta}\), we find \(tan\theta = 1⇒ \theta = 45^\circ\). Range at \(y = 0\) is \(x = 80\text{ m}\). Maximum height is \(H = \frac{Rtan\theta}{4} = 20\text{ m}\), and the initial speed \(u = 20\sqrt{2}\text{ m/s}\). Hence, A-Q, B-P, C-R, D-S.

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