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Yuliya22 [10]
2 years ago
6

An object is dropped from an airplane and takes 72 seconds to hit the ground. What is the magnitude of its VF Y ?

Physics
1 answer:
Dafna1 [17]2 years ago
3 0

The final vertical velocity of the object is 705.6 m

Explanation:

Here we have to analyze the vertical motion of the object, which is a free fall motion, with constant acceleration of

g=9.8 m/s^2

towards the ground due to gravity.

Therefore, we can use the following suvat equation:

v=u+at

where

v is the final vertical velocity

u is the initial vertical velocity

a=g is the acceleration

t is the time

Since the object is dropped,

u = 0

Therefore, the magnitude of the vertical velocity of the object after t = 72 seconds is:

v=0+(9.8)(72)=705.6 m/s

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve
Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

s_2 = gt^2/2

s_1 = g(t + \Delta t)^2/2

v_2 = gt

v_1 = g(t + \Delta t)

Their difference in distance are therefore:

\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2

\Delta s = g/2((t + \Delta t)^2 - t^2)

\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t) (AsA^2 - B^2 = (A-B)(A+B)

\Delta s = g\Delta t/2(2t + \Delta t)

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

\Delta v = v_1 - v_2 = g(t + \Delta t) - gt

\Delta v = gt + g\Delta t - gt = g\Delta t

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

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3 years ago
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PLZ HELP!!!!
noname [10]
I believe the answer would be B) ...... sorry if i'm wrong
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Answer:

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