A stunt woman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The speed of the horse is 13.5 m/s, and the woman is initially 2.60 m above the level of the saddle.

(a) What must be the horizontal distance between the saddle and limb when the woman makes her move?
m

(b) How long is she in the air?
s

a. h = Vo*t + 0.5g*t^2.= 2.60 m.

0 + 4.9t^2 = 2.60,
t^2 = 0.531,
Tf = 0.728 s = Fall time.
d = V*t = 13.5m/s * 0.728 s. = 9.83 m.

b. Tf = 0.728 s = Fall tm3e.

To find the horizontal distance, we can use the formula:

horizontal distance = horse's speed × time

To find the time, we can use the vertical motion equation:
vertical distance = initial vertical velocity × time + 0.5 × acceleration × time^2

The initial vertical velocity is 0 since the woman is stationary in the vertical direction before jumping. The acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).

(a) To find the horizontal distance, we need to find the time first.
From the vertical motion equation:
vertical distance = 0.5 × (-9.8 m/s^2) × time^2 + initial vertical velocity × time + initial vertical distance

Since the woman is initially 2.6 m above the level of the saddle:
0 = 0.5 × (-9.8 m/s^2) × time^2 + 0 × time + 2.6 m

Simplifying the equation:
4.9 × time^2 = 2.6
time^2 = 2.6 / 4.9
time ≈ √0.5306
time ≈ 0.73 s

Now, let's find the horizontal distance:
horizontal distance = 13.5 m/s × time
horizontal distance ≈ 13.5 m/s × 0.73 s
horizontal distance ≈ 9.855 m

Therefore, the horizontal distance between the saddle and limb should be approximately 9.855 meters.

(b) To find the time the woman is in the air, we can use the vertical motion equation:
vertical distance = 0.5 × (-9.8 m/s^2) × time^2 + 0 × time + initial vertical distance

Since the vertical distance is 0 (she lands on the horse):
0 = 0.5 × (-9.8 m/s^2) × time^2 + 0 × time + 2.6 m

Simplifying the equation:
4.9 × time^2 = 2.6
time^2 = 2.6 / 4.9
time ≈ √0.5306
time ≈ 0.73 s

Therefore, the woman is in the air for approximately 0.73 seconds.

To solve this problem, we need to consider the horizontal and vertical motion of the stunt woman.

Let's start with part (a) and find the horizontal distance between the saddle and limb when the woman makes her move.

We can use the formula for horizontal distance traveled (d) in uniform motion:
d = v * t

In this case, the horse's speed (v) is given as 13.5 m/s. Since the stunt woman is dropping vertically, her horizontal velocity will be the same as the horse's speed, so we can use the same value for v.

We want to find the time (t) it takes for the woman to reach the horse's level, so we'll solve for t in another equation. The vertical motion of the stunt woman can be described using the equation:
d = v_0 * t + (1/2) * a * t^2

Where:
- d is the vertical distance traveled,
- v_0 is the initial vertical velocity (in this case, 0 because she starts from rest),
- a is the acceleration due to gravity (approximately -9.8 m/s^2),
- t is the time.

In this case, the vertical distance traveled is 2.60 m. Solving the equation for t, we get:
2.60 = 0 * t + (1/2) * (-9.8) * t^2

Simplifying the equation, we have:
2.60 = -4.9t^2

Dividing both sides by -4.9, we get:
t^2 = -2.60 / -4.9

Taking the square root of both sides, we find:
t ≈ 0.63 s

Now that we have the time (t), we can find the horizontal distance (d) using the formula:
d = v * t

Plugging in the values, we have:
d = 13.5 * 0.63 = 8.505 m

So, the horizontal distance between the saddle and limb when the woman makes her move is approximately 8.505 meters.

Moving on to part (b), we need to find how long the stunt woman is in the air.

Since the horizontal motion does not affect the time in the air, we can use the same value of t that we calculated earlier, which is approximately 0.63 seconds.

Therefore, the stunt woman is in the air for approximately 0.63 seconds.

To summarize:
(a) The horizontal distance between the saddle and limb when the woman makes her move is approximately 8.505 meters.
(b) The stunt woman is in the air for approximately 0.63 seconds.