How Do You Calculate the Time and Velocity of a Dropped Object in Physics?

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SUMMARY

The discussion focuses on calculating the time and velocity of a dropped object, specifically a cell phone dropped from a hang glider descending at 8 m/s from an altitude of 30 m. The calculations reveal that the cell phone takes approximately 1.77 seconds to reach the ground, with a final velocity of -25.53 m/s just before impact. The negative sign indicates the direction of the velocity, which can be simplified by assuming downward as positive. The equations used include Vf^2 = V0^2 + 2ay and t = (vf - vi) / a.

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Homework Statement


A man using a hang glider who is descending at a speed of 8 m/s drops his cell phone from an altitude of 30 m.
a. How long does it take the cell phone to reach the ground?
b. What is the velocity of the cell phone just before it hits the ground?

Homework Equations


Vf^2=V0^2+2ay
t=v-v0/a

The Attempt at a Solution


(-8m/s)^2+2(-9.8 m/s)(-30)

I want to make sure I am using negative velocity and accel due to descending speed...
 
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64 m/s^2+(-19.6m/s)(-30m)
64+588
vf2=652
sqrt(652)
25.53
Due to descending, it is -25.53 m/s

t=vf-vi/-9.8 m/s^2

=-17.3/-9.8=1.765
t=1.77 seconds
 
a.k said:
64 m/s^2+(-19.6m/s)(-30m)
64+588
vf2=652
sqrt(652)
25.53
Due to descending, it is -25.53 m/s

t=vf-vi/-9.8 m/s^2

=-17.3/-9.8=1.765
t=1.77 seconds
those answers look good. You can avoid confusion with the minus sign by assuming down direction as positive, but what you did was fine, although you did have to have the insight that the sq root of 652 was the negative root
 

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