Answer:
Given equation of the parabola is,

where h(x) is the height function and x denotes the horizontal distance.
To find the distance thrown does the ball hit the ground.
we get that,

To find the value of x, we get

In order to cancel the denominater multiply by -20, we get

Factorise the above equation, we get
![x=\frac{160\pm\sqrt[]{(160)^2+4(120)}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/6ps2o3f5l2upt6se6e5qaigr3s3itat1ez.png)
![x=\frac{160\pm\sqrt[]{4(6400)+4(120)}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/lzy9zd1neaebbwc1mdvepbx1xqv3qm1do3.png)
![x=\frac{160\pm2\sqrt[]{6400+120}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/8fp2knpbi20lfg398x8f5ngqoikdjslh5q.png)
![x=80\pm\sqrt[]{6520}](https://img.qammunity.org/2023/formulas/mathematics/college/5cfd8d4bc7na7avr210e1oyu8cfyohvvhh.png)
![x=80\pm2\sqrt[]{1630}](https://img.qammunity.org/2023/formulas/mathematics/college/9eop830p8whmb1yyg7bfoaq35b8hezgg6k.png)
The distance between the initial and final point is,
![d=\lvert80-2\sqrt[]{1630}-(80+2\sqrt[]{1630})\rvert](https://img.qammunity.org/2023/formulas/mathematics/college/899aum0cdgje8zbadd259fx562eqc8nzag.png)
![d=4\sqrt[]{1630}](https://img.qammunity.org/2023/formulas/mathematics/college/n2ysvi18b5mqu6jodkwp0cfu0d2lq2lj9a.png)
Answer is:
![d=4\sqrt[]{1630}\approx161.49](https://img.qammunity.org/2023/formulas/mathematics/college/lf42hc5vmqikg3bwzmbrorzk2bvc0bw3u6.png)