A daring stunt woman sitting on a tree limb

wishes to drop vertically onto a horse galloping under the tree. The constant speed of the
horse is 12.5 m/s, and the woman is initially
1.31 m above the level of the saddle.
How long is she in the air? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of s

and the second question is

What must be the horizontal distance between the saddle and limb when the woman
makes her move?
Answer in units of m

See the solution in Related Questions below

To find out how long the stunt woman is in the air, we can use the kinematic equation for vertical motion:

h = v0*t + (1/2)*a*t^2

Where:
h is the initial vertical displacement (1.31 m)
v0 is the initial vertical velocity (0 m/s, as the woman starts from rest)
a is the acceleration due to gravity (-9.8 m/s^2, as it acts downward)
t is the time the woman is in the air (unknown)

Substituting the given values, the equation becomes:

1.31 = 0*t + (1/2)*(-9.8)*t^2

Simplifying further:

1.31 = -4.9t^2

Rearranging the equation:

4.9t^2 = -1.31

t^2 = -1.31 / 4.9

t = sqrt(-1.31 / 4.9)

Since the value under the square root is negative, it means there is no real solution. This implies that the stunt woman cannot drop vertically onto the horse.

For the second question, we need to find the horizontal distance between the saddle and the limb when the woman makes her move. Since the woman wants to drop onto a horse with a constant speed, we can assume that the horizontal distance covered by the horse and the stunt woman is the same.

We can use the equation of motion to find the horizontal distance:

d = v*t

Where:
d is the horizontal distance
v is the horizontal velocity (12.5 m/s, as given)
t is the time the woman is in the air (unknown)

Since we don't have the value of t, we cannot determine the exact horizontal distance without additional information or assumptions.