Projectile Motion: Glider Release Speed and Time to Reach the Ground

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Homework Help Overview

The problem involves a glider released from an airplane, initially traveling horizontally at 81 m/s, and subject to a constant acceleration of 2.4 m/s² at an angle of 1.1° below the horizontal due to air drag. The objective is to determine the time it takes for the glider to reach the ground from a height of 5.7 km.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relevance of horizontal components of acceleration and velocity to the vertical travel time. One suggests using the kinematic equation for vertical motion to solve for time.
  • Questions arise regarding how to determine the initial acceleration in the vertical direction, particularly in relation to the forces acting on the glider.
  • Another participant elaborates on the forces involved, including gravity and air drag, and how to set up the equations to find the resultant vertical acceleration.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem, including the setup of equations and the effects of forces. Some guidance has been provided regarding the use of kinematic equations, but no consensus or resolution has been reached yet.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available and the methods they can use. The specific values for forces and angles are also under discussion, impacting the calculations for vertical acceleration.

Delta G
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Homework Statement



A glider is tugged by an airplane at 81 m/s when it is released. If the original speed was along the horizontal and the glider is now under a constant acceleration of 2.4 m/s2 at 1.1° below the horizontal due to air drag, how long will it take to reach the ground 5.7 km below?

a. 250,000 s
b. 500s
c. 4.8 s
d. 2.2s

Homework Equations



v0x = v0*cos(theta)
v0y = v0*sin(theta)
vy = v0y + at
x = x0 + v0x*t
y = y0 + v0y*t + 1/2*at^2
vy^2 = v0^2 +2a(delta y)

The Attempt at a Solution




No idea on how to start.
 
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Notice that the horizontal components of acceleration and velocity do not affect the time it takes to travel vertically - the time to reach the ground.

y = y0 + v0y*t + 1/2*at^2

This equation should work. You know y, y0, v0, and a. Solve for t.
 
How do I find the initial acceleration in the y direction?
 
Delta G said:

Homework Statement



A glider is tugged by an airplane at 81 m/s when it is released. If the original speed was along the horizontal and the glider is now under a constant acceleration of 2.4 m/s2 at 1.1° below the horizontal due to air drag, how long will it take to reach the ground 5.7 km below?

a. 250,000 s
b. 500s
c. 4.8 s
d. 2.2s

Homework Equations



v0x = v0*cos(theta)
v0y = v0*sin(theta)
vy = v0y + at
x = x0 + v0x*t
y = y0 + v0y*t + 1/2*at^2
vy^2 = v0^2 +2a(delta y)

The Attempt at a Solution




No idea on how to start.

The force of air drag (F_{drag}) on the glider is in the opposite direction of the velocity of the glider. Since this drag is 1.1 degree below the horizontal, the F_{drag} will have a horizontal and vertical component (i.e. will pull the glider backward and downward). So to find the acceleration of the glider in the y direction, we find the resultant force acting on the glider, which is the sum of the y components of the forces acting on it (e.g. F_{grav} and F_{drag}).

F_{drag} = 2.4 m/s2 at 1.1° below the horizontal
y_{i} = 5.7 km = 5700 m
v_{i} = 81 m/s

\sumF = ma
\sumF_{y} = F_{grav} + F_{drag} = ma_{y}
\sumF_{y} = -mg - F_{drag}sin(1.1) = ma_{y}
a_{y} = -(g + (F_{drag}sin(1.1))/m)

From the y component of acceleration, you can derive the y equation as xcvxcvvc said, set that equation equal to zero since y = 0 at ground, and solve for t.
 

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