Package dropped from plane at 500 m/s from 2000 m altitude

[panpanthepirate]

a plane flying horizontally at 500 m/s releases a package at an altitude of 2000 meters. how long will the package take to reach ground? how far will the package travel horizontally while falling?

 

[mohlam12]

here is a hint: you have to know that the time that it is going to take for the package to go horizontally is the same to go vertically... also, when the package is going horixontally, there is no gravity...

 

[xanthym]

a plane flying horizontally at 500 m/s releases a package at an altitude of 2000 meters. how long will the package take to reach ground? how far will the package travel horizontally while falling?

{Horizontal Distance} = d = v_x0*t = (500 m/sec)*t
{Vertical Height} = h = h₀ + v_y0*t - (1/2)*g*t² =
= (2000 meters) + (0)*t - (1/2)*(9.81 m/sec²)*t² =
= (2000) - (4.91)*t²

The package will continue falling until it hits ground at time "t" given by:
h = 0 = (2000) - (4.91)*t²
::: ⇒ t² = (2000)/(4.91) = (407.3)
::: ⇒ t = (20.2 sec)

The horizontal distance "d" traveled during this time t=(20.2 sec) is thus given by:
d = (500 m/sec)*t = (500 m/sec)*(20.2 sec)
d = (10100 meters)


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[panpanthepirate]

thanks all for your help, i got another problem that says a box having a weight of 490 Newtons is dragged across the floor by means of a rope that makes an angle 30 degrees with the floor, the coeffecient of sliding friction is .300. find the force that must be applied to the rope to provid uniform velocity after the starting friction has been overcome.

 

[HallsofIvy]

1. Find the friction force: coefficient of friction times weight

2. To slide with uniform velocity, the horizontal force must equal that

3. The force applied to the rope is along the hypotenuse of a right triangle having the horizontal force as a leg- use trigonometry.

 

[EUROPE1]

{Horizontal Distance} = d = v_x0*t = (500 m/sec)*t
{Vertical Height} = h = h₀ + v_y0*t - (1/2)*g*t² =
= (2000 meters) + (0)*t - (1/2)*(9.81 m/sec²)*t² =
= (2000) - (4.91)*t²

The package will continue falling until it hits ground at time "t" given by:
h = 0 = (2000) - (4.91)*t²
::: ⇒ t² = (2000)/(4.91) = (407.3)
::: ⇒ t = (20.2 sec)

The horizontal distance "d" traveled during this time t=(20.2 sec) is thus given by:
d = (500 m/sec)*t = (500 m/sec)*(20.2 sec)
d = (10100 meters)


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Come on, write mathematical formulas:

$$t = \sqrt{\frac{2h}{g}}$$


$$D = v_xt = v\sqrt{\frac{2h}{g}}$$