In physics, it is established that the acceleration due to gravity, g (in meters divided by sec squared) in meters/sec², at a height h meters above sea level is given by the formula below. where 6.374 x 10⁶⁶ is the radius of Earth in meters.
A certain building is 458 meters tall. What is the acceleration due to gravity at the top at the top of the building? g(h) = m/sec².

Explanation:

The given function is

Now h = the height from the surface of the Earth

Here the building is 458 m tall

So,

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