Question-176146

Question Number 176146 by peter frank last updated on 13/Sep/22

Answered by mr W last updated on 13/Sep/22

Commented by mr W last updated on 13/Sep/22

b = \frac{{g t}^{2}}{2} \Rightarrow t = \sqrt{\frac{2 b}{g}}

u_{x} = \frac{a}{2 t} = \frac{a}{2} \sqrt{\frac{g}{2 b}}

u_{y} = g t = \sqrt{2 b g}

y = k x (a - x)

b = k \times \frac{a}{2} \times (a - \frac{a}{2}) \Rightarrow k = \frac{4 b}{a^{2}}

\Rightarrow y = \frac{4 b x (a - x)}{a^{2}}

Commented by peter frank last updated on 14/Sep/22

thank you

Commented by Tawa11 last updated on 15/Sep/22

Great sir

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