Time for Specific Velocity Direction in Projectile – Rankers Physics
Topic: Kinematics
Subtopic: Ground to Ground Projectile

Time for Specific Velocity Direction in Projectile

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.
\(1\text{ sec}\)
\(3\text{ sec}\)
\(5\text{ sec}\)
\(7\text{ sec}\)

Solution:

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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