Ground to Ground Projectile: Practice Problem & Solution
A particle is projected from horizontal ground with speed \(50\text{ ms}^{-1}\) at \(53^\circ\) with horizontal. Find time after which velocity of particle will be at \(45^\circ\) with horizontal for the second time.
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:
Horizontal speed is \(u_x = u\cos(53^\circ) = 30\text{ m/s}\), and vertical speed is \(u_y = u\sin(53^\circ) = 40\text{ m/s}\). At \(45^\circ\) on the way down, \(v_y = -v_x = -30\text{ m/s}\). Using \(v_y = u_y - gt\), we get \(-30 = 40 - 10t\), which yields \(t = 7\text{ s}\).
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