PLEASE HELP WILL MARK AS BRAINLIEST Drag each label to the correct location.
Match the chemical equations to the appropriate types of chemical reactions.
H2CO3-H20 + CO2
8Cu + Sg - 8 Cus
BaCO3-BaO + CO2
24+502- 2P205
KCIO3-KCI + O2
8Fe + S8 - 8Fes

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

higher potential

explanation:

because it is in a high speed so it must move into a of high potential