Cars a and b are racing each other along the same straight road in the following manner: car a has a head start and is a distance da beyond the starting line at t=0. the starting line is at x=0. car a travels at a constant speed va. car b starts at the starting line but has a better engine than car a, and thus car b travels at a constant speed vb, which is greater than va. how far from car b's starting line will the cars be when car b passes car a?

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

a. carbon monoxide from car exhaust

b.particulate matter, such as soot

c.water vapor from a factory smokestack

d.acid rain

these are the answer choices im on that question right now