Question

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.)

Answer 1

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.

Answer 2

106 ft/s.

I had this question too, all I did was convert the other answer to feet lol.