The velocity of the skydiver after 1 second is -9.8 m/s
Explanation:
To solve this problem, we just need to consider the vertical motion of the skydiver.
The skydiver is in free fall, it means that it is moving with a uniformly accelerated motion, with constant acceleration [tex]g=9.8 m/s^2[/tex] towards the ground. Therefore, its velocity at time t is given by the following suvat equation:
[tex]v=u-gt[/tex]
where
v is the velocity
u is the initial velocity
g is the acceleration of gravity
t is the time
Here we have:
u = 0, because initially the skydiver has no velocity
[tex]g=9.8 m/s^2[/tex]
And substituting t = 1 s, we find the skydiver's velocity after 1 second:
[tex]v=0-(9.8)(1)=-9.8 m/s[/tex]
Where the negative sign means the direction is downward.
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