When a particle is projected vertically upwards with an init
When a particle is projected vertically upwards with an initial velocity of v_0 = 198 m/s, it experiences an acceleration a = -(9.81 + 0.014V^2), where v is the velocity of the particle. Determine the maximum height reached by the particle. h_Max =
Solution
as v2=u2+2gh where v =final velocity , u= initial velocity, g is gravitational force/acceleration, h= max height attained
v= 0 as at heighest point speed/velocity will become zero
u=198 m /s
put values in formule:
0=198x198+2x{-(9.81+0.014v2)}Hmax
198x198/2=(9.81+0.014x198x198)Hmax
Hmax= 35.69m
