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.

Answer 1

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

Answer 2

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