A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with a speed of 6.0 m/s. What is the frog's speed as it flies off the top of the ramp?

First, let's find the angle of inclination using the tangent function.

sin θ = opposite/hypotenuse = 1 m/2 m

θ = 30°

Assuming the ramp is frictionless, the force balance is:

F = mgsinθ = ma

Cancelling out m,

a = gsinθ = (9.81 m/s²) (sin 30°) = 4.905 m/s²

Using the equation for rectilinear motion at constant acceleration,

x = v₀t + 0.5at²

2 m = (6 m/s) (t) + 0.5 (4.905 m/s²) (t²)

Solving for tm

t = 0.297 seconds

Using the equation for acceleration:

a = (v - v₀) / t

4.905 m/s² = (v - 6 m/s) / 0.297 s

Solving for v,

v = 7.46 m/s