A-particle-is-projected-upwards-with-a-velocity-of-96ms-1-In-addition-to-being-subject-to-gravity-it-is-acted-on-by-a-retardation-of-16t-where-t-is-the-time-from-the-start-of-the-motion-

Question Number 152492 by nadovic last updated on 28/Aug/21

A particle is projected upwards with

a velocity of 96 {m s}^{- 1} . In addition to

being subject to gravity, it is acted on

by a retardation of 16 t, where t is the

time from the start of the motion .

What is the greatest height attained

by the particle ?

Answered by mr W last updated on 29/Aug/21

\frac{d v}{d t} = - (g + 16 t)

\int_{v_{0}}^{v} d v = - \int_{0}^{t} (g + 16 t) d t

v = v_{0} - (g t + 8 t^{2})

0 = v_{0} - ({g t}_{1} + 8 t_{1}^{2})

8 t_{1}^{2} + {g t}_{1} - v_{0} = 0

\Rightarrow t_{1} = \frac{- g + \sqrt{g^{2} + 32 v_{0}}}{16}

\frac{d s}{d t} = v_{0} - (g t + 8 t^{2})

s = \int_{0}^{t} [v_{0} - (g t + 8 t^{2})] d t

s = v_{0} t - \frac{{g t}^{2}}{2} - \frac{8 t^{3}}{3}

\Rightarrow s_{m a x} = v_{0} t_{1} - \frac{{g t}_{1}^{2}}{2} - \frac{8 t_{1}^{3}}{3}

Commented by nadovic last updated on 29/Aug/21

Thank you Sir

Commented by SANOGO last updated on 29/Aug/21

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