Asmall 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

░░░░░░░░░░░░▄▐░░░░░░ ░░░░░░▄▄▄░░▄██▄░░░░░ ░░░░░▐▀█▀▌░░░░▀█▄░░░ ░░░░░▐█▄█▌░░░░░░▀█▄░ ░░░░░░▀▄▀░░░▄▄▄▄▄▀▀░ ░░░░▄▄▄██▀▀▀▀░░░░░░░ ░░░█▀▄▄▄█░▀▀░░░░░░░░ ░░░▌░▄▄▄▐▌▀▀▀░░░░░░░ ▄░▐░░░▄▄░█░▀▀░░░░░░░ u have been spooked by the ▀█▌░░░▄░▀█▀░▀░░░░░░░ ░░░░░░░▄▄▐▌▄▄░░░░░░░ ░░░░░░░▀███▀█░▄░░░░░ ░░░░░░▐▌▀▄▀▄▀▐▄░░░░░ spooky skilenton ░░░░░▐▀░░░░░░▐▌░░░░░ ░░░░░░█░░░░░░░░█░░░░ ░░░░░▐▌░░░░░░░░░█░░░ ░░░░░█░░░░░░░░░░▐▌░░ send this to 4 ppl or skelintons will eat you

Average growth population growth