A skydiver deploys his parachute when he is 950m directly above his desired landing spot. He then falls through the air at a steady 5.0 m/s . There is a breeze blowing to the west at 1.1 m/s.
At what angle with respect to vertical does he fall?
By what distance will he miss his desired landing spot?
Yoe = 364(66)
To find the angle with respect to vertical at which the skydiver falls, we can use trigonometry. Let's assume that the vertical direction is positive upward.
The skydiver is falling at a steady speed of 5.0 m/s, and there is a breeze blowing to the west at 1.1 m/s. Using these velocities, we can find the resultant velocity of the skydiver's motion.
The horizontal component of velocity due to the breeze is 1.1 m/s to the west, and the vertical component of velocity is 5.0 m/s downward.
To find the resultant velocity, we use the Pythagorean theorem:
Resultant velocity = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
Plugging in the values:
Resultant velocity = sqrt((1.1)^2 + (-5.0)^2)
Resultant velocity = sqrt(1.21 + 25)
Resultant velocity = sqrt(26.21)
Resultant velocity ≈ 5.120 m/s
Now, we can find the angle with respect to vertical using the inverse tangent function:
angle = arctan(vertical velocity/horizontal velocity)
Plugging in the values:
angle = arctan(-5.0/1.1)
angle ≈ -77.64 degrees
Therefore, the skydiver falls at an angle of approximately -77.64 degrees with respect to vertical.
To find the distance by which the skydiver will miss his desired landing spot, we can calculate the horizontal displacement.
The skydiver falls for a certain amount of time, and during this time, the horizontal displacement is given by the product of the horizontal velocity and the time.
The time can be calculated using the formula: time = distance/vertical velocity
where distance is the initial vertical distance from the desired landing spot (950m) and vertical velocity is the vertical component of the resultant velocity (5.0 m/s).
Plugging in the values:
time = 950/5.0
time = 190 seconds
Now, we can calculate the horizontal displacement:
horizontal displacement = horizontal velocity * time
horizontal displacement = 1.1 * 190
horizontal displacement = 209 meters
Therefore, the skydiver will miss his desired landing spot by approximately 209 meters to the west.
To find the angle at which the skydiver falls, we can use trigonometry. The vertical component of the skydiver's velocity is the speed at which he falls, which is 5.0 m/s. The horizontal component of the skydiver's velocity is the speed of the breeze, which is 1.1 m/s.
The tangent of the angle with respect to vertical is equal to the vertical velocity component divided by the horizontal velocity component. So, we have:
tan(angle) = (vertical velocity) / (horizontal velocity)
tan(angle) = 5.0 m/s / 1.1 m/s
angle = arctan(5.0 / 1.1)
Using a calculator, we find that the angle is approximately 78.12 degrees with respect to vertical.
To calculate the distance the skydiver will miss his desired landing spot, we need to find the horizontal distance he travels during his descent.
The time it takes for the skydiver to deploy the parachute and fall through the air can be found by dividing the vertical distance above the landing spot (950 m) by the vertical velocity of the skydiver (5.0 m/s). So, we have:
time = vertical distance / vertical velocity
time = 950 m / 5.0 m/s
time = 190 seconds
During this time, the skydiver will also move horizontally due to the breeze. The distance he travels in the horizontal direction can be calculated by multiplying the time by the horizontal velocity (1.1 m/s). So, we have:
horizontal distance = time * horizontal velocity
horizontal distance = 190 s * 1.1 m/s
horizontal distance = 209 m
Therefore, the skydiver would miss his desired landing spot by approximately 209 meters.
a. Vo = 5m/s @ 270o + 1.1m/s @ 180o.
X = 5*cos270 + 1.1*cos180 = -1.1 m/s.
Y = 5*sin270 + 1.1*sin180 = -5 m/s.
tanAr = Y/X = -5/-1.1 = 4.54545
Ar = 77.6o = Reference angle.
A=180 + Ar = 180 + 77.6 = 257.6o, CCW.
A = 270 - 257.6 = 12.4o West of South.
b. Vo = -1.1/cos257.6=5.12 m/s @ 257.6o
Xo = 5.12*cos257.6 = -1.1 m/s.
Yo = 5.12*sin257.6 = -5.0 m/s.
h = Vo*t + 0.5g*t^2 = 950 m.
-5t + 4.9t^2 = 950
4.9t^2 - 5t -950 = 0
Use Quadratic formula:
Tf = 14.4 s. = Fall time.
Dx = Xo * Tf = -1.1m/s * 14.4=1.1 m.=
1.1m West of desired landing spot.