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. What is the greatest height attained by the particle?


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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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