Projectile(no height) and Displacement(Gravity)

[Evolution17]

Hello guys! First post here on the forums and I'd greatly appreciate it if I could receive some help with 2 physics questions. The first one is a projectile question, and the second involves and object falling in a certain time frame through a certain length.

1. A car is racing towards a 20 degree inclined ramp at 30m/s. The driver has to TRY and clear a piranha tank and land on an identical ramp on the opposite side.

To get vertical component I did

cos70=x/30
x=10.26m/s
***That is the vertical speed/component

And to get the horizontal speed/component I did Pythagoras theorem, I ended up getting 28.2m/s.

I know then you have to sub it into the quadratic equation but I am never given the height of the ramp, other than knowing that I will need to land on an identical ramp on the other side. So do I just sub in 0 for the "c" part? So

-10.26 +/- SQR -10.26^-4(-4.9)(0)/2(-4.9)

and then get time and then sub it into the horizontal speed/component and get the distance covered?

Second question

An object falls past a 1.9m window in 0.2s. Determine the height of the object above the window.
I know this is wrong but what I tried to do was determine the velocity (9.5m/s) and I'm GUESSING that the distance above the window is 9.5m?

Help would be greatly appreciated.

 

[SteamKing]

For problem 2, you are studying physics, not guesstimating. Use your knowledge of accelerated motion to figure the problem.

You are given the height of the window and the time it takes for the object to pass. Assume that the object is traveling at v1 when it reaches the top of the window. Since it is accelerating due to gravity, you can find the change in velocity as a function of time as the object passes the window. Having determined the velocity function, you can find out what the final velocity of the object must be as it drops below the bottom of the window. Working back and finding v1, you can then calculate how high above the window the object must have been dropped (with v0 = 0 of course).

 

[haruspex]

do I just sub in 0 for the "c" part? So

-10.26 +/- SQR -10.26^-4(-4.9)(0)/2(-4.9)

Yes, but there's something wrong in the equation. A missing 2 perhaps.

An object falls past a 1.9m window in 0.2s. Determine the height of the object above the window.

Suppose the answer is x. How long did it take to reach the top of the window? ... to reach the bottom of the window?