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The first man jumps in his parachute Does not open. He accelerates at 10 m/s squared for four seconds before the parachute opens how fast is he going just before the parachute opens

Respuesta :

skyluke89

The velocity of the parachutist after 4 seconds is 40 m/s (downward)

Explanation:

Before his parachute opens, the man is in free fall, so we can calculate its velocity at time t using the suvat equation:

v = u + at

where

v is the velocity at time t

u is the initial velocity

[tex]a=g=10 m/s^2[/tex] is the acceleration of gravity (taking downward as positive direction)

Assuming he jumps from rest,

u = 0

So, its velocity after t = 4 s is

[tex]v=0+(10)(4)=40 m/s[/tex]

And the direction is downward, since the result is positive.

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