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

[Delta G]

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.

 

[xcvxcvvc]

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.

 

[Delta G]

How do I find the initial acceleration in the y direction?

 

[GRB 080319B]

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

$$\sum$$F = ma
$$\sum$$$$F_{y}$$ = $$F_{grav}$$ + $$F_{drag}$$ = m$$a_{y}$$
$$\sum$$$$F_{y}$$ = -mg - $$F_{drag}$$sin(1.1) = m$$a_{y}$$
$$a_{y}$$ = -(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.