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 13.5 m/s, and the man is initially 3.45 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
To find the horizontal distance between the saddle and the limb, as well as the time the stuntman is in the air, we can use the equation of motion for vertical motion and the relationship between horizontal and vertical motion.
(a) To find the horizontal distance (x) between the saddle and limb, we can use the relationship between horizontal and vertical motion. When the stuntman drops vertically from the tree limb onto the horse, the horizontal distance covered by the horse is the same as the distance traveled by the stuntman in the air. Therefore, the horizontal distance is equal to the horizontal component of the horse's velocity (Vx) multiplied by the time in the air (t).
We need to find Vx, which is the horizontal component of the horse's velocity. Since only the vertical component of the velocity changes due to gravity, the horizontal component remains constant. Thus, Vx = 13.5 m/s.
Next, we need to find the time in the air (t). The vertical distance the stuntman needs to travel is the sum of the initial height above the saddle (3.45 m) and the height of the saddle. Since we're assuming the saddle's height is 0, the total vertical distance the stuntman needs to cover is 3.45 m.
Using the equation of motion for vertical motion:
Δy = V₀y * t + (1/2) * a * t²
Where:
Δy = change in vertical position (3.45 m)
V₀y = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s²)
t = time in the air (unknown)
Rearranging the equation, we get:
3.45 m = (1/2) * (-9.8 m/s²) * t²
Simplifying:
6.9 m = -4.9 m/s² * t²
Solving for t², we get:
t² = 6.9 m / 4.9 m/s² = 1.408 s²
Taking the square root of both sides, we find:
t ≈ 1.188 s
Now that we have the time in the air, we can find the horizontal distance (x) using the equation x = Vx * t:
x = 13.5 m/s * 1.188 s ≈ 16.02 m
Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 16.02 m.
(b) The time the stuntman is in the air is approximately 1.188 seconds.