Physics: Abasketball is shot from an initial height of 2.4m with an initial velocity of 12m/s directed at an angle 35° above the horizontal.a. how far from the basket was the player when the short was made? at what angle to the horizontal did the ball enter the basket?

Abasketball is shot from an initial height of

Ответ:

romeojose2005

12.03.2022

The distance of the player from the basket when the shot was made is 12.78 m and the angle to the horizontal that the ball entered the basket is 30.8⁰.

Your question is not complete, it seems to be missing the following information;

the height of the basket from the ground = 3.05 m

The given parameters;

initial height of the ball, h = 2.4 minitial velocity of the projection, u = 12 m/sangle of projection, Ф = 35°

The vertical distance traveled by the basketball is calculated as;

$\Delta s = 3.05 \ m - 2.4 \ m = 0.65 \ m$

The time of motion of the ball from the initial position is calculate as;

$h = v_0_yt - \frac{1}{2} gt^2\\\\ 0.65 = (12 \times cos(35))t\ - \ (0.5\times 9.8)t^2\\\\ 0.65 = 6.888t - 4.9t^2\\\\4.9t^2 - 6.888t + 0.65 = 0\\\\a = 4.9, \ b = -6.888, \ c = 0.65\\\\t = \frac{-b \ \ + /-\ \ \sqrt{b^2 -4ac} }{2a} \\\\t = \frac{-(-6.888) \ \ + /-\ \ \sqrt{(-6.888)^2 -4(4.9\times -0.65)} }{2(4.9)} \\\\t = 1.3 \ s$

The horizontal distance traveled by the ball is calculated as;

$X = v_o_xt\\\\X = (v_0 \times cos (\theta)) \times t\\\\X = (12 \times cos (35)) \times 1.3 \\\\X = 12.78 \ m$

The angle to the horizontal that the ball entered the basket is calculated as;

the vertical component of the final velocity;

$v_y_f = v_o - gt\\\\v_y_f = (12 \times sin(35) ) - (9.8 \times 1.3)\\\\v_y_f = -5.85 \ m/s$

the horizontal component of the final velocity is constant;

$v_0_x = v_f_x = 12 \times cos(35) = 9.83 \ m/s$

$tan(\theta) = \frac{v_f_y}{v_f_x} = \frac{5.85}{9.83} = 0.595\\\\\theta = tan^{-1} (0.595)\\\\\theta = 30.8^0$

Thus, the distance of the player from the basket when the shot was made is 12.78 m and the angle to the horizontal that the ball entered the basket is 30.8⁰

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Ответ:

dontcareanyonemo

12.03.2022

Physics

We'd need to know how high the basket was to get an exact answer. I'll denote it by . The ball's horizontal and vertical positions, and , respectively, are given by

$x=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\cos35^\circ t$

$y=2.4\,\mathrm m+\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\sin35^\circ t-\dfrac g2t^2$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity.

The ball reaches the basket when y=h ; we solve this equation for (use the quadratic formula) to find it does so when

$t=0.702+0.102\sqrt{94.4-19.6h}$

For example, if the net follows the NBA standard of a height of 10 ft (about 3.0 m), then we find $t=1.3\,\mathrm s$ .

At this point, we have a horizontal position of

$x=6.90+1.00\sqrt{94.4-19.6h}$

Assuming the NBA standard again, this amounts to the player taking the shot about 13 meters away from the net.

The ball's final velocity vector has components

$v_x=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\cos35^\circ$

$v_y=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\sin35^\circ-gt$

so that at the time we found above, the velocity vector has components

$v_x=9.8\,\dfrac{\mathrm m}{\mathrm s}$

$v_y=\left(0.023-1.00\sqrt{94.4-19.6h\right)\,\dfrac{\mathrm m}{\mathrm s}$

The angle the ball makes with the horizontal is then computed by solving for $\theta$ in $\tan\theta=\dfrac{v_y}{v_x}$ . Symbolically, the solution for $\theta$ will be a bit messy. If we take h=3.0 as in the above examples, we end up with