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Question Number 121180 by TITA last updated on 05/Nov/20

Commented by TITA last updated on 05/Nov/20

$$pleasehelp$$

Answered by TANMAY PANACEA last updated on 05/Nov/20

$$y=0.7x-0.02x^2$$$$y=0$$$$0.7x-0.02x^2=0$$$$x(0.7-0.02x)=0$$$$Thestonehitatadistance=\frac{0.7}{0.02}=\frac{7\times 100}{10\times 2}=35meterfrom$$$$pointO$$$$\left(vcos\theta \right)t=x$$$$y=\left(vsin\theta \right)t-\frac{1}{2}{gt}^2$$$$y=\left(\frac{vsin\theta }{vcos\theta }\right)x-\frac{1}{2}g\times \frac{x^2}{v^2{cos}^2\theta }$$$$y=\left(tan\theta \right)x-\left(\frac{1}{2}\times g\times \frac{1}{v^2{cos}^2\theta }\right)x^2$$$$comparing$$$$tan\theta =0.7\to \theta ={tan}^{-1}(0.7)$$$$\frac{1}{2}\times 10\times \frac{1}{v^2}\times {sec}^2\theta =0.02$$$$v^2=\frac{5(1+0.49)}{0.02}=\frac{1.49\times 5}{0.02}$$$$v=\sqrt{\frac{5\times 1.49}{0.02}}$$$$0^2={\left(vsin\theta \right)}^2-2gH$$$$H=\frac{v^2{sin}^2\theta }{2g}$$$$H=\frac{1.49\times 5}{0.02\times 2\times 10}\times {\left(\frac{0.7}{\sqrt{1+0.49}}\right)}^2$$$$H=\frac{5\times 0.49}{0.02\times 20}$$

Commented by TITA last updated on 05/Nov/20

$$thanksir$$

Commented by TANMAY PANACEA last updated on 06/Nov/20

$$mostwelcome$$

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