A stone thrown up from the top of a building with a velocity of 48 meter per second reaches the ground with a velocity of -50 m/s. Find the height of the building. Using the derivative of approximation.

you should know that using the metric system,

a = -4.9 m/s^2
distance = -4.9t^2 + 48t + c, where c is the initial height
velocity = -9.8t + 48
given: v = -50
-50 = -9.8t + 48
9.8t = 98
t = 10

When t=0 , distance = 0+0+c = c
when t = 10, distance = -4.9(100) + 48(10) + c
= c - 10

height of building
= difference in distances
= c - (x-10)
= 10 m

check:
dist= -4.9t^2 + 48t + 10
v = -9.8t + 48
when t = 10
v = -9.8(10) + 48 = -50 , good!
when t = 0 , distance = 0+0+10 = 10, good
when t = 10, distance = -4.9(100) + 48(10) + 10
= 0, good

a = -9.8 m/s^2 = g

(1/2) a = -4.9 m/s^2
so
distance = (1/2)g t^2 +48 t + c
distance = -4.9t^2 + 48t + c, where c is the initial height

To solve this problem using the derivative of approximation, we need to make the assumption that the only force acting on the stone is gravity, and neglect air resistance. This means that the acceleration due to gravity (-9.8 m/s^2) is the only force affecting the stone's motion.

Let's denote the initial velocity of the stone as u (48 m/s), the final velocity as v (-50 m/s), and the height of the building as h.

We can use the kinematic equation to relate the initial and final velocities, acceleration, and displacement:

v^2 = u^2 + 2as

Since the stone is thrown upwards, the acceleration due to gravity should be considered negative, so we have:

v^2 = u^2 - 2g(h)

Substituting the given values:

(-50)^2 = (48)^2 - 2(9.8)(h)

Simplifying:

2500 = 2304 - 19.6h

Rearranging the equation:

19.6h = 2304 - 2500

19.6h = -196

Dividing by 19.6, we get:

h = -196 / 19.6

h ≈ -10

However, we have to consider that the height cannot be negative. In this case, the negative sign indicates that the stone is falling downwards, but we are interested in the height of the building.

So, we discard the negative sign and take the absolute value:

h ≈ |-10|

Therefore, the height of the building is approximately 10 meters.

Note: The derivative of approximation method is not specifically required to solve this problem. It is typically used in calculus to estimate values of functions using derivatives. In this case, we used the kinematic equation to solve the problem without directly applying derivatives.