91Ó°ÊÓ

The initial speed of the shot is $12.0\text{m}/\text{s}$, the launch angle of the shot is ${40.0}^{°}$angle from the horizontal, the shot leaves the hand of the thrower at a height of $1.80\text{m}$above the ground.

The period of travel by applying kinematics relation is

$h_o=\left(u\sin \mathrm{θ}\right)t−\frac{1}{2}g{t}^2$

Here, $h_o$is the initial height of the shot, and $g$is the acceleration due to gravity.

$\begin{array}{l}(−1.8\text{m})=(12.0\text{m}/\text{s})\sin \left({40}^{{}^{°}}\right)t−\frac{1}{2}\left(9.80\text{m}/{\text{s}}^2\right)t^2\\ 4.9t^2−7.71t−1.8=0\end{array}$

Calculating the roots

$\begin{array}{l}t=\frac{−(−7.71)Â\pm \sqrt{{(−7.71)}^2−4\left(4.9\right)(−1.8)}}{2\left(4.9\right)}\\ t=\frac{7.71Â\pm 9.73}{9.8}\\ t=1.78\text{sor}t=−0.206\text{s}\end{array}$

Since, the period of travel can't be negative, so the period of travel is$1.78\text{s}$

The horizontal distance travelled by the shot is

$x=\left(u\cos \mathrm{θ}\right)t\left(\text{I}\right)$

Here, x is the horizontal distance traveled by the shot, u is the initial velocity of the shot, is the launch angle, and is the period of travel.

$x=\left[(12.0\text{m}/\text{s})\cos \left({40}^{{}^{°}}\right)\right](1.78\text{s})=16.4\text{m}$

Conclusion:

Therefore, the distance travelled by the shot is$16.4\text{m}$