A ball leaves a table with an initial horizontal velocity of 2.4m/s. The height of the table is adjustable. In two runs, the height of the table is set at 2.5m and 5.0m respectively. Assume air resistance is negligible. What is the horizontal distance between the two landing locations of the ball?
Xo = 2.4 m/s
h = 0.5g*t^2 = 2.5 m.
4.9*t^2 = 2.5
t^2 = 0.510
Tf = 0.714 s. = Fall time.
D1 = Xo * Tf = 2.4m/s * 0.714s = 1.71 m.
4.9*t^2 = 5 m.
t^2 = 1.02
Tf = 1.01 s.
D2 = 2.4m/s * 1.01s = 2.42 m.
D2-D1 = 2.42 - 1.71 = 0.714 m.