Motion of a block on a steep surface

[diredragon]

Homework Statement

The block was given an initial velocity up the surface with an angle of 45 degrees to the ground. Calculate the $\frac{t_1}{t_2}$ with $t_1$ being the time it took to get to the highest point up the hill and $t_2$ the time it took to get down. In both cases the frictional force with coefficient m= 0.2 was acting on it.
2. Homework Equations
3. The Attempt at a Solution
My tried solution is uploaded below in form of a picture.
I have made free body diagrams but am stuck at getting the equation that involves time so that i can divide the two times. Any hints?
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[PeroK]

I have made free body diagrams but am stuck at getting the equation that involves time so that i can divide the two times. Any hints?

Think generally: if a body moves a distance $d$ how does the time depend on the acceleration?

 

[diredragon]

Think generally: if a body moves a distance $d$ how does the time depend on the acceleration?

$d=v_ot + \frac{1}{2}at^2$
Im still kinda stuck...

 

[PeroK]

$d=v_ot + \frac{1}{2}at^2$
Im still kinda stuck...

The body goes the same distance up the slope as it does down the slope. Down the slope is easy, as $v_o = 0$.

You can use a trick for motion up the slope: the time taken to move a distance $d$ from an initial velocity to rest at deceleration $a$ is the same as the time to move a distance $d$ from rest at acceleration $a$. This avoids needing to calculating $v_0$.

To see this, think about distance as the area under a velocity/time graph.

 

[diredragon]

The body goes the same distance up the slope as it does down the slope. Down the slope is easy, as $v_o = 0$.

You can use a trick for motion up the slope: the time taken to move a distance $d$ from an initial velocity to rest at deceleration $a$ is the same as the time to move a distance $d$ from rest at acceleration $a$. This avoids needing to calculating $v_0$.

To see this, think about distance as the area under a velocity/time graph.

I don't understand the trick, so in the new equation for the motion up the slope i would not have $v_o$ but would have some new $a$ while d stays the same right? How would i make that work?