A parachutist bails out and freely falls 59 m. Then the parachute opens, and thereafter she decelerates at 1.6 m/s2. She reaches the ground with a speed of 2.5 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?
To find the time the parachutist is in the air, we can divide the problem into two parts: the free fall and the deceleration phase.
(a) To find the time during the free fall phase, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity = 0 m/s (since the parachutist stops at the end of free fall)
u = initial velocity (unknown)
a = acceleration due to gravity = 9.8 m/s^2
s = distance traveled during free fall = 59 m
Rearranging the equation, we have:
u^2 = v^2 - 2as
u^2 = 0 - 2 * 9.8 * 59
u^2 = - 2 * 9.8 * 59
Since the initial velocity cannot be negative, we take the positive square root:
u = √(- 2 * 9.8 * 59)
Now, we can use the equation:
v = u + at
Where:
v = final velocity = 2.5 m/s
u = initial velocity (which we just found)
a = deceleration due to parachute = -1.6 m/s^2 (negative because it opposes the direction of motion)
t = time during deceleration phase (unknown)
Rearranging the equation, we have:
t = (v - u) / a
t = (2.5 - u) / -1.6
Substituting the value of u, we can solve for t:
t = (2.5 - √(- 2 * 9.8 * 59)) / -1.6
(b) To find the height at which the fall begins, we need to consider the total distance traveled during the free fall and deceleration phases. This can be calculated by finding the area under the velocity-time graph:
Distance = Area of the triangle + Area of the rectangle
For the triangle, the base is the time spent in free fall (which we found in part a), and the height is the initial velocity during free fall:
Area of the triangle = (1/2) * (u) * (t)
For the rectangle, the base is the time spent during the deceleration phase, and the height is the final velocity during free fall:
Area of the rectangle = (v) * (t)
Adding both areas gives us the total distance traveled:
Distance = (1/2) * (u) * (t) + (v) * (t)
Substituting the values of u and t, we can find the height:
Distance = (1/2) * (√(- 2 * 9.8 * 59)) * [(2.5 - √(- 2 * 9.8 * 59)) / -1.6] + (2.5) * [(2.5 - √(- 2 * 9.8 * 59)) / -1.6]
Now, we can calculate both the time spent in the air (part a) and the height at the beginning of the fall (part b).
To solve this problem, let's break it down into two parts: the initial free fall and the subsequent deceleration with the parachute.
(a) To find the time the parachutist is in the air, we can use the following equation:
v = u + at
First, let's calculate the initial velocity (u) during the free fall. Since the parachutist is freely falling, u = 0 m/s.
The final velocity (v) during the free fall is given as 2.5 m/s. The acceleration (a) during the free fall is due to gravity, which is approximately 9.8 m/s² in the downward direction.
Using the equation v = u + at, we can rearrange it to solve for time (t):
t = (v - u) / a
Plugging in the values, we have:
t = (2.5 m/s - 0 m/s) / 9.8 m/s²
Calculating this, we get:
t = 0.2551 seconds
Therefore, the parachutist is in the air for approximately 0.2551 seconds.
(b) To find the height at which the fall begins, we can use the equation for distance covered during free fall:
s = ut + 0.5at²
Here, the initial velocity (u) is 0 m/s, acceleration (a) is approximately 9.8 m/s², and time (t) is 0.2551 seconds.
Plugging the values in, we have:
s = 0 * 0.2551s + 0.5 * 9.8 m/s² * (0.2551 s)²
Calculating this, we get:
s = 0.3085 meters
Therefore, the fall begins at a height of approximately 0.3085 meters.