Answer:
It leaves the table traveling at 2.35 m/s and falls for 0.639 s.
Explanation:
This is a projectile motion problem. I will use the vertical motion to find the time in flight first, then I will calculate the launch speed. I think it is easiest this way.
Like any object that falls 2.00 m, the time required comes from
Δ
y
=
v
o
Δ
t
+
1
2
a
(
Δ
t
)
2
In this case,
v
o
=
0
,
Δ
y
=
−
2.00
m
(because it is falling) and
a
=
−
9.8
m
s
23
−
2.00
=
−
1
2
(
9.8
)
(
Δ
t
)
2
(
Δ
t
)
2
=
2.00
4.9
=
0.408
Δ
t
=
0.639
s
To travel 1.50 m in 0.639 s, the initial speed must have been
v
o
=
1.50
m
0.639
s
=
2.35
m
s
