A ball is launched horizontally from the edge of a table. It strikes the floor.
ID: 2220551 • Letter: A
Question
A ball is launched horizontally from the edge of a table. It strikes the floor. The horizontal distance between the point on the floor directly beneath the edge of the table and the point on the floor where the ball strikes is 3.23 m. The vertical distance between the floor and the edge of the table is 1.17 m. Find (a) the time interval from when the ball was launched to when the ball struck the floor and (b) the initial speed of the ball. Next, the ball is launched from the floor at an angle 55.7 degrees above the horizontal with the same initial speed. It again strikes the floor. Find (c) the time interval from when the ball was launched to when the ball strikes the floor and (d) the horizontal distance from the point where the ball was launched to the point where the ball strikes the floor.Explanation / Answer
Dy = [(-½) • g • t²] + [Vyø • t]
where Vyø = 0 = initial velocity in the y_direction,
and Dy = the y_position as a function of time.
-1.17 = [(-½) • (9.8) • t²]
t = 0.494 sec = fall time
Dx = Vxø • t
3.23 = Vxø • (0.495)
Vxø = 6.57 m/sec
______________________________________…
Vø = 6.57 m/sec @ +55.7º
Vxø = (6.57) • cos[55.7] = 3.48 m/sec
Vyø = (6.57) • sin[55.7] = 5.57 m/sec
Vy = [-g • t] + Vyø
Vy = 0 at the maximum altitude
0 = [(-9.8) • t] + 5.57
t = 0.568 sec
T = 2 • t = time the ball is in the air since uptime = downtime.
T = 1.14 sec
Dx = Vxø • T
Dx = (3.48) • (1.14)
Dx = 3.96 m
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