VLR help me with physics

3. Keanu hopes that there is some "incline" that will assist them. Assume that the opposite side of the gap is 1 meter lower than the takeoff point. Also, the stunt drivers that launch this bus clearly have the assistance of a "takeoff ramp" from which the bus launches at an angle. Assume that the ramp is angled at 3.00° above the horizontal. Prove whether or not the bus will make it to the opposite side. 4. What is the minimum angle necessary for the bus to launch and make the opposite side of the gap?
   Can someone answer #4 for me, I already solved #3

I'm still stuck on 56 x 6 so I can't help him

i just copy and pasted your post into google and this popped up.

http://faculty.polytechnic.org/physics/3%20A.P.%20PHYSICS%202009-2010/02.%202-d%20motion%20(kin,%20cent%20acc,%20rel%20motn)/2._pdf's/ch4Lect_white.pdf

Your question is on slide 16, I don't know if the answer is on there since it aint my homework

keanu is a fucking dumbass

i think u should use the formula of s= ut-1/2gt^2 and then try to find theta in range formula

maybe im dumb that i dont understand this question

Do we know the length of the gap?

too stupid for this

The bus will not make it across the gap with a launch angle of 3.00 ∘ and an initial velocity of 30 m/s.
The minimum angle required for the bus to make it across the gap is approximately 6.3°.

Given: h = 1m, d = 15.24m, v_0 = 29.95 m/s. Use trig to find v_x0 =29.91 m/s and v_y0 = 1.57 m/s.
We need to make the simplification that if the front of the bus "reaches" the other end, they will make it, as otherwise we need to know things about the bus (what if the front tires land but nothing else does...) We also ignore air resistance because I'm assuming that this is basic physics.

We need to calculate how far the bus will travel in the positive x direction in the time that it takes to go up, reach the apex, fall to its original height, and then fall one meter. we use y = y0 + vy_0t + 0.5at^2, or, y = 0 + 1.57t - 4.9t^2 => t=0.639s. Since we have v_x0 (and we assume that this velocity is constant), the same formula for x gives us x = x0 + v_x0 + 0.5at^2 =0+29.91t + 0.5 x 0 x t^2 = 29.91t. We have t, so the bus will travel 19.11 meters by the time it falls the full one meter. Since the gap was given to be 15.24, they make it!

You mentioned that you only needed question four, but the exact same framework can be used, just in the reverse direction. The minimum theta for which this works is when v_x0t = 15.24. v_x0 = 29.95cos(theta). Then, 29.95cos(theta) x t = 15.24. Isolating for t, t = 0.509/cos(theta). Similarly, we find that v_y0 = 29.95sin(theta). Great! Just like last time, we use y = y0 + vy_0t + 0.5at^2, or, y = 29.95sin(theta)t - 4.9t^2. Now, we substitute in our expression for t. y = 29.95sin(theta) x 0.509/cos(theta) - 4.9(0.509/cos(theta))^2. Cleaning up, we get y = 15.24tan(theta) - 1.27/cos(theta)^2. We set this to negative one and solve for a theta in the first quadrant. I found theta = 0.01774 radians = 1.0164 degrees!

I am certain that this solution is far more convoluted than it needs to be (and it might even be wrong) but I hope it helps! Feel free to reply with questions or concerns. (This also took me like 10 minutes...)

Unrelated, is this from Speed? I swear I did this same assignment last year...