(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a\(32^\circ \)ramp at a speed of\(40.0{\rm{ m}}/{\rm{s}}\)\(\left( {144{\rm{ km}}/{\rm{h}}} \right)\). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are\(20.0{\rm{ m}}\)long?

(b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

When a projectile is launched at an angle to the ground with a certain beginning velocity, it follows a curved parabolic path. The horizontal range is the distance between the point of projection and the place of landing.

• Angle of the inclination of the ramp \(\theta = 32^\circ \).
• Speed of the motorcycle \(u = 40.0\,{\rm{m}}/{\rm{s}}\).
• Length of the bus \(L = 20.0\,{\rm{m}}\)

The horizontal range covered by the daredevil is,

\(R = \frac{{{u^2}\sin \left( {2\theta } \right)}}{g}\)

Here\(u\) is the initial velocity,\(\theta \) the angle of projection or angle of the ramp, and\(g\) is the acceleration due to gravity.

Substitute\(40{\rm{ m}}/{\rm{s}}\) for\(u\),\(32^\circ \) for\(\theta \), and\(9.8{\rm{ m}}/{{\rm{s}}^2}\) for\(g\),

\[\begin{array}{c}R = \frac{{{{\left( {40{\rm{ m}}/{\rm{s}}} \right)}^2} \times \sin \left( {2 \times 32^\circ } \right)}}{{\left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 146.74{\rm{ m}}\end{array}\]

The number of buses safely cleared by the daredevil is,

\(n = \frac{R}{L}\)

Here\(R\) is the horizontal range and\(L\) is the length of one bus.

Substitute\(146.74{\rm{ m}}\) for\(R\), and\(20.0{\rm{ m}}\) for\(L\),

\(\begin{array}{c}n = \frac{{146.74{\rm{ m}}}}{{20.0{\rm{ m}}}}\\ = 7.34\end{array}\)

Hence, the daredevil will safely clear the \(7\) buses.

The total length of the\(7\) buses is,

\({R_L} = 7 \times L\)

Here\(L\) is the length of one bus.

Substitute\(20.0{\rm{ m}}\) for\(L\),

\(\begin{array}{c}{R_L} = 7 \times \left( {20{\rm{ m}}} \right)\\ = 140{\rm{ m}}\end{array}\)

The margin of error is,

\(\gamma = \frac{{R - {R_L}}}{R} \times 100\% \)

Here\(R\) is the horizontal range and\({R_L}\) is the total length of\(7\) buses.

Substitute\(146.74{\rm{ m}}\) for\(R\), and\(140{\rm{ m}}\) for\({R_L}\),

\(\begin{array}{c}\gamma = \frac{{\left( {146.74{\rm{ m}}} \right) - \left( {140{\rm{ m}}} \right)}}{{\left( {146.74{\rm{ m}}} \right)}} \times 100\% \\ = 4.6\% \end{array}\)

Hence, the margin of error is \(4.6\% \). This is not enough, especially considering that air resistance was neglected by our calculation.