Question

A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into the air and onto a track that is velocity $2.0m$ lower.
The cyclist travels a horizontal distance of $6.0m$ in the air. Air resistance is negligible. What is the horizontal velocity $v$ of the cyclist at the end of the light track?

• A. $6.3m{s}^{-1}$
• B. $9.4m{s}^{-1}$
• C. $9.9m{s}^{-1}$
• D. $15m{s}^{-1}$

Answer 1

The correct option is B $9.4m{s}^{-1}$

Consider the problem,

This is a projectile motion question So, the horizontal and vertical component of the motion can be considered separately. The quantity that can relate both motion is the time $t$.

Initially, since the cyclist is on a horizontal track he has only a horizontal velocity of $v$, and no vertical velocity.

Air resistance is negligible, so the only force (causing an acceleration) that effects the motion is the force of gravity (which is downwards).

Consider the vertical component

Initial velocity, $u=0$

$\begin{array}{c}Acceleration,a=9.81m{s}^{-2}\left(duetogravity\right)\\ Dis\tan ce,s=2.0m\\ Time,t=?\left(timetoreachthelowerhorizontaltrack\right)\end{array}$

$\begin{array}{c}s=ut+\frac{1}{2}a{t}^2\\ 2=0+\frac{1}{2}\times 9.81\times t^2\\ t=\sqrt{\frac{2\times 2}{9.81}}=0.64s\end{array}$

So, The cyclist takes $0.64s$ to fall a vertical distance of $2.0m$. During this $0.64s$ he also moves a horizontal distance of $6.0m$ simultaneously.

Consider the Horizontal component

$\begin{array}{c}Acceleration,a=0\\ initialspeed=finalspeed=V\\ Dis\tan cetravelled=6.0m\\ Timetaken=0.64s\end{array}$

$\begin{array}{c}SpeedV=\frac{dis\tan ce}{time}\\ =\frac{6.0}{0.64}\\ =9.4m{s}^{-1}\end{array}$

Hence, Option $B$ is the correct answer.