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Question Number 8667 by tawakalitu last updated on 20/Oct/16

Answered by sandy_suhendra last updated on 20/Oct/16

at t=5 sec the body reaches the ground ⇒ y=0 y=V_o .t − (1/2)g.t^2 0=V_o .5 − (1/2).10.5^2 5V_o =125 ⇒ V_o = 25 m/s at t=4 sec the body reaches poin A so y_A =V_o .t_A −(1/2)g.t_A ^2 =25.4 − (1/2).10.4^2 =100−80 = 20 m

$$\mathrm{at}\:\mathrm{t}=\mathrm{5}\:\mathrm{sec}\:\mathrm{the}\:\mathrm{body}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{ground}\:\Rightarrow\:\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{V}_{\mathrm{o}} .\mathrm{t}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}.\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{V}_{\mathrm{o}} .\mathrm{5}\:−\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{10}.\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{5V}_{\mathrm{o}} =\mathrm{125}\:\Rightarrow\:\mathrm{V}_{\mathrm{o}} =\:\mathrm{25}\:\mathrm{m}/\mathrm{s} \\ $$$$\mathrm{at}\:\mathrm{t}=\mathrm{4}\:\mathrm{sec}\:\mathrm{the}\:\mathrm{body}\:\mathrm{reaches}\:\mathrm{poin}\:\mathrm{A} \\ $$$$\mathrm{so}\:\:\:\:\mathrm{y}_{\mathrm{A}} =\mathrm{V}_{\mathrm{o}} .\mathrm{t}_{\mathrm{A}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}.\mathrm{t}_{\mathrm{A}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{25}.\mathrm{4}\:−\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{10}.\mathrm{4}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{100}−\mathrm{80}\:=\:\mathrm{20}\:\mathrm{m} \\ $$

Commented by tawakalitu last updated on 20/Oct/16

Thank you sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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