When a particle is projected vertically upwards with an init

When a particle is projected vertically upwards with an initial velocity of v_o = 162 m/s, it experiences an acceleration a = -(9.81 + 0.016V^2), where v is the velocity of the particle. Determine the maximum height reached by the particle. h_Max =

Solution

Given

v₀ = 162 m/s

acceleration of the particle is governed by the equation

a = - ( 9.81 + 0.016 v² )

we know that

v² - v₀² = 2 * a * H

At maximum height v = 0

0 - 162² = 2 * [ - ( 9.81 + 0.016 * 0² ) ] * H_max

H_max = 1337.61 meters