Aweight of mass m1 is on
africtionless ramp which makes angle theta with respect to
thehorizontal. this mass starts at the top of the ramp which is
atheight l above the horizontal. if the mass is simply
leftto slide down the ramp, how long will it take to reach
thehorizontal?

Explanation:

The mass will have a weight, and since it is on a surface it will have a normal reaction.

The vertical component of the normal reaction will be equal and opposite to the weight.

w = g * m

Nv = N * sin(θ)

N is the normal reaction and Nh its vertical component

Nv = -w

N * sin(θ) = -g * m

The horizontal component of the normal will be

Nh = N * cos(θ)

N = Nh / cos(θ)

Then:

Nh / cos(θ) * sin(θ) = -g * m

sin/cos = tg

Nh * tg(θ) = -g * m

The horizontal component of the normal force will be the only force in the horizontal direction

It will cause an acceleration

Nh = ah * m

Then

ah * m * tg(θ) = -g * m

Simplifying the mass on each side

ah * tg(θ) = -g

ah = -g * tg(θ)

The mass will slide from a height related to the lenght of the ramp

L = D * sin(θ)

D = L / sin(θ) This is the distance it will slide

We set up a reference system with origin at the top of the ramp and the positive X axis pointing down the ramp in the direction of the slope.

In this reference system:

X(t) = X0 + V0*t + 1/2 * a * t^2

X0 = 0

V0 = 0

Then

X(t) = -1/2 * g * tg(θ) * t^2

It will move the distance D

L / sin(θ) = -1/2 * g * tg(θ) * t^2

t^2 = L / ( sin(θ) * (-1/2) * g * tg(θ) )

The negative sign will dissapear because gravity has a negative sign too.