Question

jana is standing on a hillside and jeni is standing on a hilltop. The horizontal direction of Jana and Jeni is 30 m. The slope of the hill satisfies equation y=0.4x (assuming jane and jeni are located in xy coordinates). If jana shoots a ball with an elevation angle of 50 degrees with respect to the horizontal direction, what is the speed of the ball so it exactly reaches Jeni?

Answer 1

Answer on Question #62465, Physics / Mechanics | Relativity

Jana is standing on a hillside and Jeni is standing on a hilltop. The horizontal direction of Jana and Jeni is $30\mathrm{m}$ . The slope of the hill satisfies equation $y = 0.4x$ (assuming Jane and Jeni are located in xy coordinates). If Jana shoots a ball with an elevation angle of 50 degrees with respect to the horizontal direction, what is the speed of the ball so it exactly reaches Jeni?

Solution:

$\mathrm{v_{xo} = v_o\cos 50^o = 0.643v_o}$

$\mathrm{v_{yo} = v_o\sin 50^o = 0.766v_o}$

For the hill, $\theta = \tan^{-1} 0.4 = 21.8{}^{\circ}$

$\mathrm{r_{Jeni}} = 30 \mathrm{~m}$

$\mathrm{x_{Jeni}} = (30\mathrm{m})(\cos 21.8{}^{\circ}) = (30\mathrm{m})(0.928) = 27.85\mathrm{m}$

$\mathrm{y_{Jeni}} = (30 \mathrm{~m})(\sin 21.8{}^{\circ}) = (30 \mathrm{~m})(0.371) = 11.14 \mathrm{~m}$

$\mathrm{x_{ball}} = \mathrm{v_{xo}t}$

$\mathrm{y_{ball}} = \mathrm{v_{yo}t} + (1 / 2)\mathrm{a_y}t^2 = \mathrm{v_{yo}t} - (1 / 2)\mathrm{g}t^2$

For some time $t$ , we require $x_{\text{ball}} = x_{\text{Jeni}}$ and $y_{\text{ball}} = y_{\text{Jeni}}$

$\mathrm{x_{ball}} = \mathrm{x_{Jeni}}$

$\mathrm{v_{xo}t} = 27.85 \mathrm{~m}$

$0.643\mathrm{v_o}t = 27.85\mathrm{m}$

$t = (27.85\mathrm{m}) / (0.643\mathrm{v_o})$

$t = 43.31 / v_{o}$

Now we use this value of the time in

$\mathrm{y_{ball}} = \mathrm{y_{Jeni}}$

$\mathrm{v_{yo}t - (1 / 2)g t^{2} = 11.14m}$

$[0.766\mathrm{v_o}]\cdot [43.31 / \mathrm{v_o}] - (1 / 2)\cdot [9.8]\cdot [43.31 / \mathrm{v_o}]^2 = 11.14$

33.18 - 9191 / $\mathrm{v_o^2} = 11.14$

9191 / $\mathrm{v_o^2} = 22.04$

$\mathrm{v_o^2} = 9191 / 22.04 = 417$

$\mathrm{v_o} = 20.4 \mathrm{~m} / \mathrm{s}$

Answer: $20.4 \, \text{m/s}$

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