The rider begins with

\(\displaystyle{K}{E}={\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}={0.5}{\left({120}{k}{g}\right)}{\left({22.5}\frac{{m}}{{s}}\right)}^{{2}}={30375}\ {J}\)

and

\(\displaystyle{P}{E}={m}{g}{h}={120}\cdot{9.8}\cdot{0}={0}\)

so,

\(\displaystyle{E}_{{t}}={K}{E}={30375}\ {J}\)

where

KE=kinetic energy

PE=potential energy

E=total energy=KE+PE

At the top of the hill some of the KE is turned into PE=mgh

PE becomes

\(\displaystyle{P}{E}={120}{k}{g}\cdot{9.8}\frac{{m}}{{s}^{{2}}}\cdot{11.0}{m}={12936}\ {J}\)

and since total energy is conserved KE becomes

\(\displaystyle{K}{E}={E}_{{t}}-{P}{E}={30375}\ {J}-{12936}\ {J}={17439}\ {J}\)

\(\displaystyle{K}{E}={\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}\Rightarrow{v}=\sqrt{{{\frac{{{2}\cdot{K}{E}}}{{{m}}}}}}=\sqrt{{{\frac{{{2}\cdot{17439}}}{{{120}}}}}}={17.05}\ \frac{{m}}{{s}}\)

Now since the hill is 11.0 m high the period of time of thevertical fall is given by

\(\displaystyle{h}={v}_{{0}}{t}+{\frac{{{1}}}{{{2}}}}{a}{t}^{{2}}\)

h=11.0m

\(\displaystyle{v}_{{0}}={0}\) (this is the initial vertical velocity)

a=9.8m/s

solve for t using the quadratic formula

\(\displaystyle{t}={\frac{{-{v}_{{a}}\pm\sqrt{{{{v}_{{0}}^{{2}}}-{4}{\left({\frac{{{1}}}{{{2}}}}{a}\right)}{\left(-{h}\right)}}}}}{{{2}{\left({\frac{{{1}}}{{{2}}}}{a}\right)}}}}\)

\(\displaystyle{t}=\pm{\frac{{\sqrt{{{2}{a}{h}}}}}{{{a}}}}\)

choose the positive answer, as the negative answer for time is meaningless

\(\displaystyle{t}={\frac{{\sqrt{{{2}{\left({9.8}\right)}{\left({11}\right)}}}}}{{{9.8}}}}={1.498}{s}\approx{1.5}\ {\sec{}}\)

so we know it takes 1.5 sec to hit the ground and we know the horizontal velocity is v=17.05m/s, distance travelled before impact:

\(\displaystyle{d}={v}{t}={17.05}\frac{{m}}{{s}}\ \cdot{1.5}{s}={25.5}{m}\)

\(\displaystyle{K}{E}={\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}={0.5}{\left({120}{k}{g}\right)}{\left({22.5}\frac{{m}}{{s}}\right)}^{{2}}={30375}\ {J}\)

and

\(\displaystyle{P}{E}={m}{g}{h}={120}\cdot{9.8}\cdot{0}={0}\)

so,

\(\displaystyle{E}_{{t}}={K}{E}={30375}\ {J}\)

where

KE=kinetic energy

PE=potential energy

E=total energy=KE+PE

At the top of the hill some of the KE is turned into PE=mgh

PE becomes

\(\displaystyle{P}{E}={120}{k}{g}\cdot{9.8}\frac{{m}}{{s}^{{2}}}\cdot{11.0}{m}={12936}\ {J}\)

and since total energy is conserved KE becomes

\(\displaystyle{K}{E}={E}_{{t}}-{P}{E}={30375}\ {J}-{12936}\ {J}={17439}\ {J}\)

\(\displaystyle{K}{E}={\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}\Rightarrow{v}=\sqrt{{{\frac{{{2}\cdot{K}{E}}}{{{m}}}}}}=\sqrt{{{\frac{{{2}\cdot{17439}}}{{{120}}}}}}={17.05}\ \frac{{m}}{{s}}\)

Now since the hill is 11.0 m high the period of time of thevertical fall is given by

\(\displaystyle{h}={v}_{{0}}{t}+{\frac{{{1}}}{{{2}}}}{a}{t}^{{2}}\)

h=11.0m

\(\displaystyle{v}_{{0}}={0}\) (this is the initial vertical velocity)

a=9.8m/s

solve for t using the quadratic formula

\(\displaystyle{t}={\frac{{-{v}_{{a}}\pm\sqrt{{{{v}_{{0}}^{{2}}}-{4}{\left({\frac{{{1}}}{{{2}}}}{a}\right)}{\left(-{h}\right)}}}}}{{{2}{\left({\frac{{{1}}}{{{2}}}}{a}\right)}}}}\)

\(\displaystyle{t}=\pm{\frac{{\sqrt{{{2}{a}{h}}}}}{{{a}}}}\)

choose the positive answer, as the negative answer for time is meaningless

\(\displaystyle{t}={\frac{{\sqrt{{{2}{\left({9.8}\right)}{\left({11}\right)}}}}}{{{9.8}}}}={1.498}{s}\approx{1.5}\ {\sec{}}\)

so we know it takes 1.5 sec to hit the ground and we know the horizontal velocity is v=17.05m/s, distance travelled before impact:

\(\displaystyle{d}={v}{t}={17.05}\frac{{m}}{{s}}\ \cdot{1.5}{s}={25.5}{m}\)