Solving for Time of Flight of Stunt Vehicle

[Format]

A stunt vehical leaves and incline with a speed of 35 m/s at a height of 52-m above level ground. Air reistance is negligible.

Find the vehicals time of flight?

Well i found the x and y components of the vehicals velocity. And i know to use d = Vit + 1/2at² giving me:

-52 = 16.4t + 1/2(-9.8)t²

My question is why can you do this? The distance of 52-m only includes the hieght of the cliff, not the height it reaches above the cliff due to leaving it on an incline. Why doesn't that make a difference in the time?

 

[UrbanXrisis]

the answer is that equation already takes into account the "height it reaches above the cliff due to leaving it on an incline." That's what the 16.4t is.

If the height is only 52m, then it would be d=.5at^2

 

[Format]

The 16.4 is the initial velocity though. Wouldnt the answer to -52 = 16.4t + 1/2t² be the time it takes from ground level to reach a height of 52 meters?

The total distance in the Y direction is more the 52...i don't understand lol

 

[UrbanXrisis]

-52 = 16.4t + 1/2(-9.8)t²
This is correct!

The total distance is more than 52, yes. That's why the equation is:
d=Vit+.5at^2
Notice the Vi*t, that's that accounts for the extra distance

so really, your distance is d-Vi=.5at^2
-52-16.4t = + 1/2(-9.8)t²

-52-16.4t is the real distance, see now?

 

[Format]

ah ok i think i got it. Thx

 

[UrbanXrisis]

yeah, np, that's what this physics fourm is for, and I love it!