Pool players often pride themselves on their ability to impart a large speed on
ID: 1332214 • Letter: P
Question
Pool players often pride themselves on their ability to impart a large speed on a pool ball. In the sport of billiards, event organizers often remove one of the rails on a pool table to allow players to measure the speed of their break shots (the opening shot of a game in which the player strikes a ball with his pool cue). With the rail removed, a ball can fly off the table; the distance the pool ball lands from the table indicates the initial speed of the ball. As the only participant with a physics background, they have placed you in charge of determining the speed of the players' break shots.
You would like to shoot an orange in a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. On your first try, you fire the arrow at 37.0 m/s at an angle of 30.0° above the horizontal from a height of 1.10 m while standing 60.0 m away. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has traveled the 60.0 m horizontally to the tree? At what speed did his pool ball hit the ground?
Explanation / Answer
Refer above figure,
Vi=37m/s, q = 30 deg
Let us first calculate time to travel 60m along X axis , T
Use kinematic equation,
R= Vix*T
60 = 37cos30*T => T = 1.87s
In this time arrow also travel vertically,
Let us calculate height of the arrow at time T= 1.87s
Use kinematic equation
h= Viy*t+1/2*a*t^2
h=37sin30*1.87+1/2*-9.8*1.87^2 = 17.46m
Total height = h+ h= 1.1+17.46=18.46 m
To calculate velocity when it hits ground use equation along y axis,
Vfy^2+Viy^2+2*a*h
Vfy^2= (37sin30)+2*-9.8*-1.1
Vfy = 19.07m/s
As acceleration is constant along X axis
Vfx= 37cos30 = 32.04m/s
Vf=sqrt(Vfy^2+Vfx^2) = sqrt( 32.04^2+19.07^2) = 37.3m/s