What is her speed at the bottom of the slope?
SOMEONE ASKED 👇
What is her speed at the bottom of the slope?
HERE THE ANSWERS 👇

Θ = arcsin(30/410) = 4.2º
weight downslope = mgsinΘ = 70kg 9.8m/s² sin4.2 = 50.2 N
friction upslope = 15 N
net downslope force = 35.2 N, so
net downslope acceleration = 35.2N / 70kg = 0.50 m/s²
Now use Torricelli’s equation:
v² = u² + 2as = (11m/s)² + 2 0.50m/s² 410m = 533 m²/s²
v = 23 m/s
Hope this helps!

A steady 15N drag force due to air resistance acts on her –that isn;t how it works