A skydiver jumps from his aircraft and falls towards earth. Due to the earth’s gravitational attraction, the sky diver will accelerate for a duration of 5.65 secs, at which time he reaches terminal velocity, that is a constant falling velocity of 181 ft./sec due to atmospheric resistance. Assuming he reaches this terminal velocity at a height of 10000 ft,
i Specify an equation in point slope form which yields the sky diver’s height as a function of time (seconds)..
ii Convert the equation into slope intercept form and plot the equation, height as a function of time.
iii At what point in time will the sky diver fall to a height of 1000 ft? How do we make that determination?
i) The equation in point-slope form can be written as follows:
h - h1 = m(t - t1)
Where:
h = height (in feet) as a function of time
h1 = initial height (10000 ft in this case)
m = slope (acceleration due to gravity for this problem)
t = time (in seconds)
t1 = initial time (0)
We know that the skydiver reaches terminal velocity after 5.65 seconds. So, for t <= 5.65, the acceleration is given by gravity (32.2 ft/sec^2).
Therefore, the equation becomes:
h - 10000 = 32.2t
ii) To convert the equation into slope-intercept form, we need to isolate 'h' on one side of the equation:
h = 32.2t + 10000
Now, we can plot this equation where 'h' is the y-axis and 't' is the x-axis.
iii) To determine at what point in time the skydiver falls to a height of 1000 ft, we can substitute 'h' with 1000 in the equation and solve for 't':
1000 = 32.2t + 10000
Rearranging the equation:
32.2t = -9000
t = -9000 / 32.2
t ≈ -279.50 seconds
However, negative time does not make physical sense in this context. Therefore, there is no point in time where the skydiver falls to a height of 1000 ft.
To answer these questions, we need to understand the physics of motion and the concept of terminal velocity.
i) The equation in point-slope form that yields the skydiver's height as a function of time can be derived using the formula for free fall motion:
h(t) = h0 + v0t + (1/2)gt^2
where:
- h(t) is the height of the skydiver at time t
- h0 is the initial height (given as 10000 ft)
- v0 is the initial velocity (0 ft/s as the skydiver is not moving initially)
- g is the acceleration due to gravity (32 ft/s^2)
Considering that the skydiver reaches terminal velocity after 5.65 seconds, we can modify the equation for after this point:
h(t) = h0 + terminal_velocity * (t - t_terminal) + (1/2)g(t - t_terminal)^2
where:
- terminal_velocity is the constant falling velocity after reaching terminal velocity (given as 181 ft/s)
- t_terminal is the time at which the skydiver reaches terminal velocity (given as 5.65 sec)
ii) To convert the equation to slope-intercept form, we simplify it by expanding and rearranging terms:
h(t) = h0 + terminal_velocity * t - terminal_velocity * t_terminal + (1/2)g(t^2 - 2t_terminalt + t_terminal^2)
Simplifying further:
h(t) = (-gt_terminal^2)/2 + terminal_velocity * t + [h0 + (gt_terminal^2)/2 + terminal_velocity * t_terminal]
The equation in slope-intercept form becomes:
h(t) = terminal_velocity * t + c
where c = h0 + (gt_terminal^2)/2 + terminal_velocity * t_terminal
iii) To determine at what point in time the skydiver will fall to a height of 1000 ft, we'll substitute h(t) with 1000 ft in the equation we derived in the previous step:
1000 = terminal_velocity * t + c
Rearranging the equation to solve for t:
t = (1000 - c) / terminal_velocity
Substituting the values:
- terminal_velocity = 181 ft/s
- c = h0 + (gt_terminal^2)/2 + terminal_velocity * t_terminal
- h0 = 10000 ft
- g = 32 ft/s^2
- t_terminal = 5.65 sec
Evaluating the equation will give us the time at which the skydiver falls to a height of 1000 ft.