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

AI Thread Summary
A glider, initially traveling at 81 m/s horizontally, is released and experiences a downward acceleration of 2.4 m/s² at an angle of 1.1° due to air drag. The problem involves calculating the time it takes for the glider to reach the ground from a height of 5.7 km. The vertical motion is governed by the equation y = y0 + v0y*t + 1/2*at², where the initial vertical velocity must be determined from the angle of release. The discussion emphasizes that only the vertical components of motion affect the time to reach the ground, and the correct approach involves analyzing forces to find the resultant vertical acceleration. The solution requires setting the vertical position equation to zero and solving for time, leading to a final answer.
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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