Question

a basketball 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?

Answer 1

We'd need to know how high the basket was to get an exact answer. I'll denote it by $h$. The ball's horizontal and vertical positions, $x$ and $y$, 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 $t$ (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

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

$v_y=-5.9\,\dfrac{\mathrm m}{\mathrm s}$

which gives an angle with the horizontal of about $\theta=-31^\circ$.

Answer 2

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 m
• initial velocity of the projection, u = 12 m/s
• angle 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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