Question Number 152492 by nadovic last updated on 28/Aug/21
$$\:\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{upwards}\:\mathrm{with} \\ $$$$\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\:\mathrm{96}{ms}^{−\mathrm{1}} .\:\mathrm{In}\:\mathrm{addition}\:\mathrm{to} \\ $$$$\:\mathrm{being}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{gravity},\:\mathrm{it}\:\mathrm{is}\:\mathrm{acted}\:\mathrm{on} \\ $$$$\:\mathrm{by}\:\mathrm{a}\:\mathrm{retardation}\:\mathrm{of}\:\mathrm{16}{t},\:\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{the} \\ $$$$\:\mathrm{time}\:\mathrm{from}\:\mathrm{the}\:\mathrm{start}\:\mathrm{of}\:\mathrm{the}\:\mathrm{motion}. \\ $$$$\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{attained} \\ $$$$\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle}? \\ $$
Answered by mr W last updated on 29/Aug/21
![(dv/dt)=−(g+16t) ∫_v_0 ^v dv=−∫_0 ^t (g+16t)dt v=v_0 −(gt+8t^2 ) 0=v_0 −(gt_1 +8t_1 ^2 ) 8t_1 ^2 +gt_1 −v_0 =0 ⇒t_1 =((−g+(√(g^2 +32v_0 )))/(16)) (ds/dt)=v_0 −(gt+8t^2 ) s=∫_0 ^t [v_0 −(gt+8t^2 )]dt s=v_0 t−((gt^2 )/2)−((8t^3 )/3) ⇒s_(max) =v_0 t_1 −((gt_1 ^2 )/2)−((8t_1 ^3 )/3)](https://www.tinkutara.com/question/Q152521.png)
$$\frac{{dv}}{{dt}}=−\left({g}+\mathrm{16}{t}\right) \\ $$$$\int_{{v}_{\mathrm{0}} } ^{{v}} {dv}=−\int_{\mathrm{0}} ^{{t}} \left({g}+\mathrm{16}{t}\right){dt} \\ $$$${v}={v}_{\mathrm{0}} −\left({gt}+\mathrm{8}{t}^{\mathrm{2}} \right) \\ $$$$\mathrm{0}={v}_{\mathrm{0}} −\left({gt}_{\mathrm{1}} +\mathrm{8}{t}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$$\mathrm{8}{t}_{\mathrm{1}} ^{\mathrm{2}} +{gt}_{\mathrm{1}} −{v}_{\mathrm{0}} =\mathrm{0} \\ $$$$\Rightarrow{t}_{\mathrm{1}} =\frac{−{g}+\sqrt{{g}^{\mathrm{2}} +\mathrm{32}{v}_{\mathrm{0}} }}{\mathrm{16}} \\ $$$$\frac{{ds}}{{dt}}={v}_{\mathrm{0}} −\left({gt}+\mathrm{8}{t}^{\mathrm{2}} \right) \\ $$$${s}=\int_{\mathrm{0}} ^{{t}} \left[{v}_{\mathrm{0}} −\left({gt}+\mathrm{8}{t}^{\mathrm{2}} \right)\right]{dt} \\ $$$${s}={v}_{\mathrm{0}} {t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{8}{t}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\Rightarrow{s}_{{max}} ={v}_{\mathrm{0}} {t}_{\mathrm{1}} −\frac{{gt}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{8}{t}_{\mathrm{1}} ^{\mathrm{3}} }{\mathrm{3}} \\ $$
Commented by nadovic last updated on 29/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$
Commented by SANOGO last updated on 29/Aug/21
