Answer:
The initial speed of the tiger is 1.80 m/s
Explanation:
Hi there!
The equation of the position vector of the tiger is the following:
r = (x0 + v0 · t, y0 + 1/2 · g · t²)
Where:
r = position vector at a time t.
x0 = initial horizontal position.
v0 = initial horizontal velocity.
t = time.
y0 = initial vertical position,
g = acceleration due to gravity.
Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:
r0 = (0.00, 6.00) m
We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.
y = y0 + 1/2 · g · t²
When the tiger reaches the ground, y = 0:
0 = 6.00 m - 1/2 · 9.81 m/s² · t²
2 · (-6.00 m) / -9.81 m/s² = t²
t = 1.11 s
We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:
x = x0 + v0 · t
At t = 1.11 s, x = 2.00 m
x0 = 0
2.00 m = v0 · 1.11 s
2.00 m / 1.11 s = v0
v0 = 1.80 m/s
The initial speed of the tiger is 1.80 m/s
to 
(from (-a/2) to
(a/2))