Suppose that an object is thrown down towards the ground from the top of the 1100-ft tall Sears
Tower with an initial velocity of – 20 feet per second (the negative velocity indicates that the object
is moving towards the ground). Given that the acceleration due to gravity of the object is a constant –
32 feet per second per second, determine how far above the ground the object is exactly six seconds
after being thrown.
h = Hi + Vi t + (1/2) a t^2
Hi = 1100
Vi = - 20
a = -32
so
h = 1100 -20 t - 16 t^2
if t=6
h = 1100 -20*6 - 16*36
= 404 ft
all this from integrating acceleration
a = -32 ft/s^2
so
dv/dt = -32
v = Vi - 32 t
dh/dt = v = Vi - 32 t
h = Hi + Vi t - (1/2)(32)t^2
To determine how far above the ground the object is exactly six seconds after being thrown, we need to use the kinematic equation for displacement.
The equation for displacement is:
d = ut + (1/2)at^2
Where:
d = displacement (distance traveled)
u = initial velocity
a = acceleration
t = time
Given:
u = -20 ft/s (negative velocity indicates downward motion)
a = -32 ft/s^2
t = 6 s
Plugging in the values, we have:
d = (-20)(6) + (1/2)(-32)(6)^2
Simplifying the equation:
d = -120 - 576
d = -696 ft
Therefore, the object is exactly 696 feet above the ground, with the negative sign indicating that it is below the ground level.