A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 10.0 m/s, and the man is initially 2.75 m above the level of the saddle.

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

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

for (b) he's in the air for 0.7s

s=Ut+0.5at^2
2.4=0t+0.5(9.8)(t^2)
t=0.7

To determine the horizontal distance between the saddle and limb when the man makes his move, we can use the equation for vertical motion.

The vertical distance can be found using the equation:

Δy = v0y * t + (1/2) * a * t^2

Where:
Δy is the vertical distance
v0y is the initial vertical velocity (0 m/s as he is dropped vertically)
t is the time in seconds
a is the acceleration (acceleration due to gravity = -9.8 m/s^2)

Given that the initial vertical distance (Δy) is 2.75 m, and the acceleration (a) is -9.8 m/s^2, the equation becomes:

2.75 = 0 * t + (1/2) * (-9.8) * t^2

Simplifying the equation gives:

2.75 = -4.9 * t^2

Rearranging the equation gives:

t^2 = 2.75 / -4.9

t^2 = -0.561

Since time cannot be negative, we discard the negative solution.

t = √(2.75 / 4.9)

t ≈ 0.595 s

The horizontal distance can be found using the equation:

Δx = v0x * t

Where:
Δx is the horizontal distance
v0x is the constant horizontal velocity of the horse (10 m/s)
t is the time in seconds

Given that the horizontal velocity (v0x) is 10 m/s, and the time (t) is 0.595 s, the equation becomes:

Δx = 10 * 0.595

Δx ≈ 5.95 m

Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 5.95 meters.

For part (b), the time in the air has already been calculated as 0.595 seconds.

To solve this problem, we can use the principles of projectile motion. Let's break it down into two parts:

(a) To determine the horizontal distance between the saddle and the limb, we can use the equation of motion for horizontal motion:

distance = speed * time

Since the horizontal speed of the horse is constant at 10.0 m/s, we can assume that the time it takes for the stuntman to drop vertically onto the horse is the same as the time it takes for the horse to travel the horizontal distance. Therefore, we need to calculate the time it takes for the stuntman to reach the horse.

To do this, we can use the equation of motion for vertical motion:

height = initial velocity * time + (1/2) * acceleration * time^2

Since the stuntman is dropping vertically, the initial velocity is 0 m/s, and the acceleration is due to gravity, which is approximately 9.8 m/s^2. The height is given as 2.75 m. Solving this equation for time, we get:

2.75 = 0 * time + (1/2) * 9.8 * time^2

Simplifying the equation:

4.9 * time^2 = 2.75

Dividing both sides by 4.9:

time^2 = 2.75 / 4.9

Taking the square root of both sides:

time = sqrt(2.75 / 4.9)

Now that we have the time it takes for the stuntman to reach the horse, we can calculate the horizontal distance using the equation distance = speed * time:

distance = 10.0 * sqrt(2.75 / 4.9)

Simplifying the equation:

distance = 10.0 * sqrt(0.561)

Calculating the value:

distance ≈ 7.49 m

Therefore, the horizontal distance between the saddle and the limb when the man makes his move is approximately 7.49 m.

(b) To determine how long the stuntman is in the air, we can use the equation of motion for vertical motion:

height = initial velocity * time + (1/2) * acceleration * time^2

Since the initial velocity is 0 m/s and the acceleration is due to gravity (-9.8 m/s^2), we can solve this equation for time when the height is 0 (which represents when the stuntman lands on the horse):

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

Simplifying the equation:

-4.9 * time^2 = 0

Dividing both sides by -4.9:

time^2 = 0

Taking the square root of both sides:

time = 0

Therefore, the stuntman is in the air for 0 seconds, as he drops vertically onto the horse without any horizontal displacement.

In conclusion, the horizontal distance between the saddle and the limb when the man makes his move is approximately 7.49 m, and the stuntman is in the air for 0 seconds.