A parachutist bails out and freely falls 50m. Then the parachute opens and thereafter she decelerates at 2 m/s^2. She reaches the ground with a speed of 3m/s.
a) how long is the parachutist in the air?
b)At what height does the fall begin?
for a) would I use x-x0 =vt-1/2 at^2 and find t? where x-x0= 50m, v= 3m/s, a = 9.8m/s^2
I am not sure what equation to use for b)
I would answer this in 2 parts. First, I would calculate the time she is in the air BEFORE she opens the chute...using x = v0t - 1/2 at^2. v0 here is 0, so you can calculate time. Then I would calculate the final velocity WHEN she opens the chute using
vf = v0 + at. v0 is 0, and you now know time t.
Then, knowing the velocity when the chute opens, you can calulate the time using vf = v0 - 2t.
To get the total time, just add the 2 times you got.
To get the total height...well, you already know the first part...50m.
To get the second height, use x = v0t-1/2 at^2, and add the two.
I hope this explains it. If not, repost!
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To solve part a), you can use the equation x - x0 = vt - 1/2 at^2, where x is the final position (ground level), x0 is the initial position (where she bailed out), v is the final velocity (3m/s), and a is the acceleration (-2m/s^2) while decelerating with the parachute.
However, it's important to note that the acceleration value you provided (9.8m/s^2) is the acceleration due to gravity, which is not applicable here because the parachutist is decelerating due to air resistance.
Let's calculate the time it takes for the parachutist to reach the ground:
Using x - x0 = vt - 1/2 at^2:
0 - 50 = 3t - 1/2(-2)t^2
Rearranging the equation and setting it equal to zero:
-2t^2 + 3t - 50 = 0
Using the quadratic formula (a standard method to solve quadratic equations):
t = (-b ± √(b^2 - 4ac)) / (2a)
Applying the values a = -2, b = 3, and c = -50 into the formula:
t = (-3 ± √(3^2 - 4(-2)(-50))) / (2(-2))
Calculating the square root and simplifying:
t = (-3 ± √(9 - 400)) / -4
t = (-3 ± √(-391)) / -4
Since the square root of a negative number is imaginary, this equation has no real solutions. It means that the parachutist would reach the ground in infinite time, which is not possible in reality.
This suggests that there might be a mistake in either the problem statement or the values given.
For part b), you can calculate the height from which the fall begins. Since the parachutist falls freely for 50m before the parachute opens, the height from which the fall begins would be 50m above the ground level.