A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 14.0 m/s, and the distance from the limb to the level of the saddle is 2.00 m.
(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
(b) How long is he in the air?
To solve this problem, we can use the kinematic equations of motion. Let's break down the problem into parts:
(a) Finding the horizontal distance between the saddle and limb:
When the ranch hand drops vertically from the tree limb, he will also have a horizontal velocity component, equal to the horizontal speed of the horse. Since the ranch hand and the horse are moving at the same speed, the horizontal distance they cover in the same time interval will be the same.
To find the horizontal distance, we can use the formula: distance = speed × time
Since the speed of the horse is given as 14.0 m/s, and the time is the same for both the ranch hand and the horse, we need to find the time it takes for the ranch hand to drop from the limb to the saddle level.
We can calculate the time using the equation: time = distance / speed
Since the vertical distance from the limb to the saddle level is 2.00 m, we can use this value as the distance and divide it by the speed of the horse to get the time.
Substituting the values, we have:
time = 2.00 m / 14.0 m/s
time ≈ 0.143 s
Now that we have the time, we can calculate the horizontal distance using the formula:
distance = speed × time
Substituting the values, we have:
distance = 14.0 m/s × 0.143 s
distance ≈ 2.002 m
Therefore, the horizontal distance between the saddle and limb when the ranch hand makes his move would be approximately 2.002 m.
(b) Finding the time the ranch hand is in the air:
The time the ranch hand is in the air can be determined using the vertical motion equation in which the initial vertical velocity is zero (since the ranch hand drops vertically) and the acceleration due to gravity is -9.8 m/s^2 (assuming downward as negative).
The equation for vertical displacement is: displacement = initial velocity × time + (1/2) × acceleration × time^2
Since the displacement is given as 2.00 m, and the initial velocity is zero, we can rearrange the equation to solve for time:
2.00 m = (1/2) × (-9.8 m/s^2) × time^2
time^2 = -2.00 m × (2 / (-9.8 m/s^2))
time^2 = 0.408 s^2
Taking the square root of both sides, we find:
time ≈ 0.639 s
Therefore, the ranch hand is in the air for approximately 0.639 seconds.
find the time to drop 2 m. d=1/2 g t^2
then, using that time, figure horizontaldistance= 14*time
I hope the horses head clears the man.