91Ó°ÊÓ

The origin is considered to be the shooting position of the basketball, where the upward direction is taken as positive y. First, calculate the proper initial speed of the ball, which covers a distance of$\mathbf{11}\mathbf{m}\mathbf{Â\pm }\mathbf{0}\mathbf{.}\mathbf{22}$.

Given data:

The basketball leaves the player’s hand at the height of $h=2.10\mathrm{m}$.

The height of the basket is $d=3.05\mathrm{m}$.

The angle at which the ball is shot is $\mathrm{θ}=38°$.

The shot is taken from a distance of $L=11\mathrm{m}$.

The accuracy of horizontal distance is $Â\pm 0.22\mathrm{m}$.

The relation of time is given by:

$t=\frac{\mathrm{Δ}x}{v\cos \mathrm{θ}}$

Here, $\mathrm{Δ}x$is the horizontal distance, and v is the initial speed of the ball.

The relation from the equation of motion is given by:

$y=\left(v\sin \mathrm{θ}\right)t+\frac{1}{2}g{t}^2$

Here, y is the distance whose value is equal to $y=d-h$, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

role="math" localid="1644987754907" $\begin{array}{c}y=\left(v\sin \mathrm{θ}\right)\left(\frac{\mathrm{Δ}x}{v\cos \mathrm{θ}}\right)-\frac{1}{2}g{\left(\frac{\mathrm{Δ}x}{v\cos \mathrm{θ}}\right)}^2\\ y=\left(\mathrm{Δ}x\tan \mathrm{θ}\right)-\frac{1}{2}\left(\frac{g\mathrm{Δ}{x}^2}{v^2{\cos }^2\mathrm{θ}}\right)\\ v=\sqrt{\frac{g{\left(\mathrm{Δ}x\right)}^2}{2{\cos }^2\mathrm{θ}\left(-y+\mathrm{Δ}x\tan \mathrm{θ}\right)}}…\left(i\right)\end{array}$

Substitute $\left(L-0.22\right)\mathrm{m}$for $\mathrm{Δ}x$in equation (I).

$\begin{array}{l}v=\sqrt{\frac{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right){\left(L-0.22\mathrm{m}\right)}^2}{2{\cos }^{2}38°\left(-d-h+\left(L-0.22\mathrm{m}\right)\tan 38°\right)}}\\ v=\sqrt{\frac{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right){\left(11\mathrm{m}-0.22\mathrm{m}\right)}^2}{2{\cos }^{2}38°\left(-\left(3.05\mathrm{m}-2.10\mathrm{m}\right)+\left(11\mathrm{m}-0.22\mathrm{m}\right)\tan 38°\right)}}\\ v=\sqrt{\frac{1140.004}{9.26}}\mathrm{m}/\mathrm{s}\\ v=11.09\mathrm{m}/\mathrm{s}\end{array}$

Substitute $\left(L+0.22\right)\mathrm{m}$for $\mathrm{Δ}x$in equation (I).

$\begin{array}{l}v=\sqrt{\frac{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right){\left(L+0.22\mathrm{m}\right)}^2}{2{\cos }^{2}38°\left(-d-h+\left(L+0.22\mathrm{m}\right)\tan 38°\right)}}\\ v=\sqrt{\frac{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right){\left(11\mathrm{m}+0.22\mathrm{m}\right)}^2}{2{\cos }^{2}38°\left(-\left(3.05\mathrm{m}-2.10\mathrm{m}\right)+\left(11\mathrm{m}+0.22\mathrm{m}\right)\tan 38°\right)}}\\ v=\sqrt{\frac{1234.96}{9.68}}\mathrm{m}/\mathrm{s}\\ v=11.29\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=11.09\mathrm{m}/\mathrm{s}$and $v=11.29\mathrm{m}/\mathrm{s}$form the range of the initial speeds of the ball.