h = 0.5g*t^2.
2.80 = 16t^2
t = 0.424s = Fall time.
a. d = Xo*t.
12.75 = Xo*0.424.
Xo = 30.1 ft/s = Initial hor. velocity of the ball.
Xo = 30.1ft/s * 1mi/5280ft * 3600s/h =
20.52 mi/h.
b. Y^2 = Yo^2 + 2g*h.
Y^2 = 0 + 64*2.80 = 179.2
Y = 13.40 ft./s. = Ver. component of final velocity.
V = sqrt(Xo^2+Y^2).
Xo = 30.1ft/s, Y = 13.40ft/s.
The top of the pool table is 2.80 ft from the floor. The placement of the tape is such that 0 feet aligns to the edge of the table (as shown). The winner of the competition wants to know if he has broken the world record for the break shot of 32 mph. His ball landed a distance 12.75 ft from the table edge. Calculate his break shot speed in mph and At what speed did his pool ball hit the ground?
1 answer
1 answer