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Question Number 195771 by Humble last updated on 10/Aug/23

Answered by liuxinnan last updated on 10/Aug/23

Answered by mr W last updated on 10/Aug/23

u_(0x) =6 m/s  u_(0z) =5(√3) m/s  g_x =−g sin θ  g_z =−g cos θ  x=u_(0x) t−((g sin θ )/2)t^2   z=u_(0z) t−((g cos θ )/2)t^2 =0  ⇒t=((2u_(0z) )/(g cos θ))=((2×5(√3))/(10 cos 30°))=2 seconds  x=6×2−((10×sin 30°)/2)×2^2 =2 m  that means after 2 seconds and after  2 meters the ball falls back onto the  field.

$${u}_{\mathrm{0}{x}} =\mathrm{6}\:{m}/{s} \\ $$$${u}_{\mathrm{0}{z}} =\mathrm{5}\sqrt{\mathrm{3}}\:{m}/{s} \\ $$$${g}_{{x}} =−{g}\:\mathrm{sin}\:\theta \\ $$$${g}_{{z}} =−{g}\:\mathrm{cos}\:\theta \\ $$$${x}={u}_{\mathrm{0}{x}} {t}−\frac{{g}\:\mathrm{sin}\:\theta\:}{\mathrm{2}}{t}^{\mathrm{2}} \\ $$$${z}={u}_{\mathrm{0}{z}} {t}−\frac{{g}\:\mathrm{cos}\:\theta\:}{\mathrm{2}}{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{u}_{\mathrm{0}{z}} }{{g}\:\mathrm{cos}\:\theta}=\frac{\mathrm{2}×\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{10}\:\mathrm{cos}\:\mathrm{30}°}=\mathrm{2}\:{seconds} \\ $$$${x}=\mathrm{6}×\mathrm{2}−\frac{\mathrm{10}×\mathrm{sin}\:\mathrm{30}°}{\mathrm{2}}×\mathrm{2}^{\mathrm{2}} =\mathrm{2}\:{m} \\ $$$${that}\:{means}\:{after}\:\mathrm{2}\:{seconds}\:{and}\:{after} \\ $$$$\mathrm{2}\:{meters}\:{the}\:{ball}\:{falls}\:{back}\:{onto}\:{the} \\ $$$${field}. \\ $$

Commented by Humble last updated on 10/Aug/23

Great!   Thanks sir.

$$\mathrm{Great}!\: \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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