AboutJames J. Kovalcin Expertise I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.
Divers in Acapulco dive from a cliff that is 64 m high. If the rocks below the cliff extend outward for 24 m, what is the minimum horizontal velocity a diver must have to clear the rocks?
Answer Almost all two dimensional kinematics can be done in the same way!
Uniform acceleration involves six possible variables: do, df, Vo, Vf, a and t as well as two equations: df=1/2*a*t^2+Vo*t+do and Vf=a*t+Vo. Since there are 6 variables and 2 equations, any time you can identify 4 of the 6 possible variables the problem is solved! In the case of a 2 dimensional problem such as this you will need to use this approach separately for each dimension: horizontal and vertical.
If we apply this approach to the vertical the 6 variables will be: do=64m [the height of the cliff], df=0m [the location of the rocks], Vo=0m/s [the diver begins with an initial vertical velocity of zero assuming that the diver dives horizontally], Vf=? [unknown], a=-9.8m/s^2 [for all freefalling objects in the vertical direction & t=? [unknown].
Using the two kinematics equations:
df=1/2*a*t^2+Vo*t+do and Vf=a*t+Vo
0=-4.9*t^2+0*6+64 and Vf=-9.8*t+0
Solve the first equation for time: 4.9*t^2=64 which becomes t=sqrt[64/4.9]=3.61s
Now repeat for the horizontal direction: do=0m [assume that the edge of the cliff is the origin], df=24m [the distance of the rocks from the base of the cliff, Vo=? [the answer to the problem!], Vf=Vo [the same since there will no significant acceleration in the horizontal direction], a=0 [as mentioned before] and t=3.61s [as calculated from the vertical data].
Again using the two available kinematics equations:
df=1/2*a*t^2+Vo*t+do and Vf=a*t+Vo
24=0+Vo*3.61+0 and Vf=0+Vo=7.6m/s
Solve the first equation for the initial velocity Vo: Vo=24/3.61=7.6m/s [the answer]
