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Question-180917




Question Number 180917 by anne1344 last updated on 19/Nov/22
Answered by mr W last updated on 19/Nov/22
c)  h_(max) =(((v sin θ)^2 )/(2g))=(((15×sin 20°)^2 )/(2×9.81))=1.34m  d)  t=((v sin θ)/g)=((15×sin 20°)/(9.81))=0.52 s
$$\left.{c}\right) \\ $$$${h}_{{max}} =\frac{\left({v}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\left(\mathrm{15}×\mathrm{sin}\:\mathrm{20}°\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{9}.\mathrm{81}}=\mathrm{1}.\mathrm{34}{m} \\ $$$$\left.{d}\right) \\ $$$${t}=\frac{{v}\:\mathrm{sin}\:\theta}{{g}}=\frac{\mathrm{15}×\mathrm{sin}\:\mathrm{20}°}{\mathrm{9}.\mathrm{81}}=\mathrm{0}.\mathrm{52}\:{s} \\ $$

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