Math 401 Velocity and Net Change Problem 2 (Falling Body.) A Skydiver jumps out of a plane falling(in m/s) at a velocity v(t)-9.8t. After 10 seconds her chute deploys and she instantly slows to a constant speed of 10m/s until she lands on the ground. a. How far does she fall in 30 seconds? b. If she jumped from 3km, when does she land?
Solution
a)
Initial velocity, u = 0
Acceleration in the first 10 s, a = g = 9.8 m/s 2
Using th formula, s = ut + 1/2 gt 2 , where s is the distance,
s = 0 + 0.5 x 9.8 x 10 x 10
= 490 m
Distance travelled in the next 20 s is,
distance = speed x time
= 10 x 20 = 200 m
Total distance travelled in 30 s = 490 + 200
= 690 m
b)
Time taken to reach first 490 m is 10 s [From previous part]
Time taken to reach next 3000 - 490 = 2510 is
t = 2510 / 10
= 251 s
Total time taken = 10 + 251
= 261 s
.
Beyond the EU: DORA and NIS 2 Directive's Global Impact
Math 401 Velocity and Net Change Problem 2 (Falling Body-) A Skydiver.docx
1. Math 401 Velocity and Net Change Problem 2 (Falling Body.) A Skydiver jumps out of a plane
falling(in m/s) at a velocity v(t)-9.8t. After 10 seconds her chute deploys and she instantly slows
to a constant speed of 10m/s until she lands on the ground. a. How far does she fall in 30
seconds? b. If she jumped from 3km, when does she land?
Solution
a)
Initial velocity, u = 0
Acceleration in the first 10 s, a = g = 9.8 m/s 2
Using th formula, s = ut + 1/2 gt 2
, where s is the distance,
s = 0 + 0.5 x 9.8 x 10 x 10
= 490 m
Distance travelled in the next 20 s is,
distance = speed x time
= 10 x 20 = 200 m
Total distance travelled in 30 s = 490 + 200
= 690 m
b)
Time taken to reach first 490 m is 10 s [From previous part]
Time taken to reach next 3000 - 490 = 2510 is
t = 2510 / 10
= 251 s
Total time taken = 10 + 251
= 261 s
![Math 401 Velocity and Net Change Problem 2 (Falling Body.) A Skydiver jumps out of a plane
falling(in m/s) at a velocity v(t)-9.8t. After 10 seconds her chute deploys and she instantly slows
to a constant speed of 10m/s until she lands on the ground. a. How far does she fall in 30
seconds? b. If she jumped from 3km, when does she land?
Solution
a)
Initial velocity, u = 0
Acceleration in the first 10 s, a = g = 9.8 m/s 2
Using th formula, s = ut + 1/2 gt 2
, where s is the distance,
s = 0 + 0.5 x 9.8 x 10 x 10
= 490 m
Distance travelled in the next 20 s is,
distance = speed x time
= 10 x 20 = 200 m
Total distance travelled in 30 s = 490 + 200
= 690 m
b)
Time taken to reach first 490 m is 10 s [From previous part]
Time taken to reach next 3000 - 490 = 2510 is
t = 2510 / 10
= 251 s
Total time taken = 10 + 251
= 261 s](https://image.slidesharecdn.com/math401velocityandnetchangeproblem2fallingbody-askydiver-230129064510-4365974c/85/Math-401-Velocity-and-Net-Change-Problem-2-Falling-Body-A-Skydiver-docx-1-320.jpg)
