Help this problem is driving me crazy

  • Thread starter Thread starter panpanthepirate
  • Start date Start date
AI Thread Summary
A plane flying at 500 m/s releases a package from an altitude of 2000 meters, which takes approximately 20.2 seconds to reach the ground. During this time, the package travels horizontally a distance of 10,100 meters. The time taken for the package to fall vertically is the same as the time it takes to travel horizontally. The calculations involve using the equations of motion under gravity and the horizontal velocity of the plane. Understanding these principles is crucial for solving similar physics problems effectively.
panpanthepirate
Messages
2
Reaction score
0
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?
 
Physics news on Phys.org
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...
 
panpanthepirate said:
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 = vx0*t = (500 m/sec)*t
{Vertical Height} = h = h0 + vy0*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)


~~
 
Last edited:
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.
 
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.
 
xanthym said:
{Horizontal Distance} = d = vx0*t = (500 m/sec)*t
{Vertical Height} = h = h0 + vy0*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)


~~


Come on, write mathematical formulas:

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


D = v_xt = v\sqrt{\frac{2h}{g}}
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top