Problem:

(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h) . 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 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.)

Solution:

To illustrate the problem, consider the following figure:

Part A

To determine the number of buses that the daredevil can clear, we will divide the range of the projectile path by 20 m, the length of 1 bus. That is

\text{no. of bus}=\frac{\text{Range}}{\text{bus length}}

First, we need to solve for the range.

\begin{align*}
\text{Range} & =\frac{\text{v}_{\text{o}}^2\:\sin 2\theta }{\text{g}} \\
\text{Range} & =\frac{\left(40.0\:\text{m/s}\right)^2\sin \left[2\left(32^{\circ} \right)\right]}{9.81\:\text{m/s}^2} \\
\text{Range} & =146.7\:\text{m} \\

\end{align*}

Therefore, the number of buses cleared is

\begin{align*}
\text{no. of buses} & =\frac{146.7\:\text{m}}{20\:\text{m}} \\
\text{no. of buses} & =7.34\:\text{buses} \\
\text{no. of buses} & =7\:\text{buses}

\qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

Therefore, he can only clear 7 buses. 

Part B

He clears the last bus by 6.7 m, which seems to be a large margin of error, but since we neglected air resistance, it really isn’t that much room for error.