Explanation
At $$\displaystyle 45^{\circ}, v_{x}= v_{y}$$
or $$\displaystyle u_{x}= u_{y}-gt$$
$$\displaystyle \therefore t= \dfrac{u_{y}-u_{x}}{g}$$
$$\displaystyle = \dfrac{40\left ( \sin 60^{\circ}-\sin 30^{\circ} \right )}{9.8}=1.5 \ s$$
horizontal and vertical component of velocity are same and equal to $$20 cos 45^o = 10\sqrt{2}$$
hence horizontal distance is $$x = 10\sqrt{2} t$$
and vertical distance is$$s = ut + \dfrac{1}{2}at^2 = 10\sqrt{2}t - \dfrac{1}{2}gt^2$$
therefore $$h = 10\sqrt{2}t - 5t^2 = A \times 10\sqrt{2}t -B\times 200t^2$$
$$10\sqrt{2}A = 10\sqrt{2}$$
$$A = 1$$
$$200B = 5$$
$$B = \dfrac{1}{40}$$
therefore $$A:B = \dfrac{1}{\dfrac{1}{40}} = 40:1$$
Alternatively, $$y = x( A-Bx ) =0$$ when particle reaches the ground.
Hence, $$R = \dfrac{A}{B}$$
But, $$R = \dfrac{u^2 \times \sin( 2 \theta)}{g} = \dfrac{20^2 \sin 90^0}{10}= 40$$
$$\Rightarrow \dfrac{A}{B} = 40: 1$$
For triangle $$ABC\,\colon\,\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}$$
Now $$\overrightarrow{AB}+\overrightarrow{BC}+2\overrightarrow{CA}$$
$$=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}+\overrightarrow{CA}=\overrightarrow{CA}$$
$$y=0.5x-0.04x^{2}= x( 0.5-0.04x)$$
When $$y=0, x=0\: or\: R$$
$$\therefore R = \dfrac{0.5}{0.04}=12.5$$
$$\displaystyle y=x\;tan\theta\left[1-\dfrac{x}{R}\right]\Rightarrow tan\theta =\dfrac{1}{2};R=12.5m$$
But $$\displaystyle R=\dfrac{4u^{2}sin\theta\;cos\theta}{g}$$
$$\displaystyle \Rightarrow u=\sqrt{\dfrac{Rg}{2\;sin\theta\;cos\theta}}=\sqrt{\dfrac{12.5\times 10}{2\times \dfrac{1}{\sqrt{5}}\times \dfrac{2}{\sqrt{5}}}}=12.5m/s$$
Please disable the adBlock and continue. Thank you.