A skydiver jumped out of a plane, determine distance he distance he descended after 5 seconds. Gravity is pulling him down at 10
1 answer:
The distance travelled in 5 seconds is 125 m
Explanation:
The motion of the skydiver is a free fall motion, since he is acted upon the force of gravity only. Therefore, it is a uniformly accelerated motion, with constant acceleration downward, and we can find the distance he travels by using the following suvat equation:
where
s is the distance travelled
u is the initial velocity
t is the time
a is the acceleration
In this problem, we have:
u = 0 (the skydiver jumps from rest)
(acceleration of gravity)
And substituting
t = 5.0 s
we find the distance travelled:
Learn more about free fall:
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Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m
Formula:
Solving:
Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m
Formula:
Solving:(Energy associated with this stretching)
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m
Formula:
Solving:
Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m
Formula:
Solving:(Energy associated with this stretching)