# Answer to Question #2774 in Abstract Algebra for carson

Question #2774

Suppose two gliders start at the same height, one with a glide ratio of 0.3 and one with a glide ratio of 2/7. If they glide until they hit the ground, which one will have glided the farther horizontal distance?

Expert's answer

Let u=(ux,uy) be the coordinates of the initial velocity of the first glider, and v=(vx,vy) be the coordinates of the initial velocity of the second glider.

Then by the definition glider ratio is the quotient:

ux / uy = 0.3,

vx / vy = 2/7

The move of each glider can be regarded as a sum of two moves: vertical and horizontal.

The vertical move has negative acceleration

a = -9.8 m/s^2, with initial velocity uy,

while the horizontal move has constant velocity ux.

Similarly for the move of the second glider.

Since& they hit the ground at the same moment T, starting from the same height, it follows that they come the same paths:

Sy1 = uy * T& - g T^2 /2

Sy2 = vy * T& - g T^2 /2

As& S1=S2 and t is the same, it follows that uy=vy.

Denote this number by& U.

Then& ux / U = 0.3, and thus

ux = 0.3 U,

and similarly

vx=2/7 U.

Since the horizontal moves have constant velocities, we have that

Sx1 = ux * T = 0.3 * U * T = 21/70 * U * T

Sx2 = vx * T = 2/7 * U * T = 20/70 * U * T,

so

& Sx1 > Sx2

Hence the first glider come the farther horizontal distance.

Then by the definition glider ratio is the quotient:

ux / uy = 0.3,

vx / vy = 2/7

The move of each glider can be regarded as a sum of two moves: vertical and horizontal.

The vertical move has negative acceleration

a = -9.8 m/s^2, with initial velocity uy,

while the horizontal move has constant velocity ux.

Similarly for the move of the second glider.

Since& they hit the ground at the same moment T, starting from the same height, it follows that they come the same paths:

Sy1 = uy * T& - g T^2 /2

Sy2 = vy * T& - g T^2 /2

As& S1=S2 and t is the same, it follows that uy=vy.

Denote this number by& U.

Then& ux / U = 0.3, and thus

ux = 0.3 U,

and similarly

vx=2/7 U.

Since the horizontal moves have constant velocities, we have that

Sx1 = ux * T = 0.3 * U * T = 21/70 * U * T

Sx2 = vx * T = 2/7 * U * T = 20/70 * U * T,

so

& Sx1 > Sx2

Hence the first glider come the farther horizontal distance.

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