If an object is thrown in an upward direction from the top of a building 1.6 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (disregard wind resistance. round answer to nearest whole number and do not reflect negative direction in your answer.)

In this case, the object is thrown upwards from the building. Therefore, it first achieves some height before its starts dropping.

Now, when going upwards
v^2 = u^2 - 2gs

Where,
v = final velocity
u = initial velocity
g = gravitational acceleration
s = height achieved from the top of he bulding

Using the values given;
v = 0 (comes into rest before it starts dropping)
u = 21.82 mi/h = 32 ft/s
g = 9.81 m/s^2 = 32.174 ft/s^2

Then,
0^2 = 32^2 - 2*32.174*s
32^2 = 2*32.174*s
s = (32^2)/(2*32.174) = 15.91 ft

After achieving that height, it starts to drop from rest to maximum velocity when it hits the ground.
Applying the same formula;
v^2 = u^2 + 2gs

Where;
v = velocity when it hits the ground
u = initial velocity, 0 ft/s as it starts from rest
s = 15.91+1.6*10^2 = 15.91+160 = 175.91 ft

Therefore,
v^2 = 0^2 + 2*32.174*175.91
v^2 = 11319.68
v = Sqrt (11319.68) = 106.39 ft/s ≈ 32.43 m/s moving downwards.

He can play two games. 13/4=3.253.25*2=6.5

Iassume the answer is 75 feet. inwards to outwards rather than outwards to inwards. probably be 75.hachures are stokes drawn in direction of the steepest slope. a series of short parallel lines drawn on a map to indicate topographic relief. it is actually called hachures on contour lines. the small marks in the center of the feature are called hachures.observe the picture attached