drag force:

D = 1/2 ρ v2 * CdA

and the power lost to drag:

P = Dv = 1/2 ρ v3 * CdA

assumptions:

air density: ρ = 1.225 kg/m3

running speed: 7 m/s,

ear height: 57 mm

normal ear projection from head: 20 mm

velecroed projection: 5 mm

both ears count so x2

ears are rounded so use a 0.7 exposure factor

ear/flat-plate-ish Cd 1.17

projected frontal area removed:

ΔA = 2 0.057 (0.020 - 0.005) * 0.7

ΔA = 0.0012 m2

convert that to drag area:

ΔCdA = 1.17 * 0.0012

ΔCdA = 0.0014 m²

for a child runner, unning CdA is roughly around 0.24 m2, so pinning the ears back reduces drag by:

0.0014 / 0.24 =0.006

about 0.6% less air drag

At 7 m/s, power saved is:

ΔP = 1/2 1.225 7^3 * 0.0014

ΔP = 0.3 watts

but air drag is only a small part of running effort. for sprinting, air resistance is roughly %8–13 of total energy/power cost. So a 0.6% reduction in the aero part gives only:

0.006 * 0.08 to 0.13 = 0.05% to 0.08%

less total effort

since drag also increses with v3, speed gain is smaller than that. Approximataly:

Δv/v ≈ fε / (1 + 2f)

where f is the aero fraction and ε is the drag-area reduction.

f = 0.08 to 0.13 , ε = 0.006:

Δv/v = 0.0004 - 0.0006

so at 7 m/s:

Δv = 0.003 to 0.004 m/s

at 100 m, that saves roughly: 0.005 to 0.008 seconds

realistic answer: a few milliseconds over 100 m.

the velcro's main performance advantage is psychological intimidation.