A particle of mass $m$ is projected from the origin with an initial velocity $v_0$ at an angle $ heta$ to the horizontal. If the particle's trajectory is described by the equation $y = x an heta - \frac{gx^2}{2v_0^2 \cos^2 heta}$, where $g$ is the acceleration due to gravity, find the time $t$ at which the particle reaches its maximum height.

Question

A particle of mass $m$ is projected from the origin with an initial velocity $v_0$ at an angle $	heta$ to the horizontal. If the particle's trajectory is described by the equation $y = x 	an 	heta - \frac{gx^2}{2v_0^2 \cos^2 	heta}$, where $g$ is the acceleration due to gravity, find the time $t$ at which the particle reaches its maximum height.

Accepted Answer

$ t = \frac{v_0 \sin \theta}{g} $

Post Utme Oau 2022 Physics Question 16

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