a) Newton's law of motion states that the acceleration acting on a falling body
(assuming no other forces) is �
x" = -g, where x(t) is the height of the object
above the ground, t is time, and g is the gravitational constant. Find a general
expression to describe the height of the object above the ground with respect
to time, t; i.e. find x(t).
(b) Suppose the object is dropped from a height of 100 units (so x'(0) = 0 and
x(0) = 100), how long will it take to hit the ground?
a) To find a general expression to describe the height of the object above the ground with respect to time, we can solve the differential equation x" = -g, where x(t) is the height of the object above the ground, t is time, and g is the gravitational constant.
To solve the differential equation, we can integrate twice.
First, we integrate x" = -g with respect to t to find x':
∫ x" dt = ∫ -g dt
x' = -gt + c1
Next, we integrate x' = -gt + c1 with respect to t to find x:
∫ x' dt = ∫ (-gt + c1) dt
x = -0.5 * g * t^2 + c1 * t + c2
Where c1 and c2 are constants of integration.
b) In this case, the object is dropped from a height of 100 units, so x(0) = 100. This gives us the initial condition for the height of the object.
Substituting the initial condition into the expression for x(t):
100 = -0.5 * g * 0^2 + c1 * 0 + c2
100 = c2
Therefore, c2 = 100.
Now we need to find the value of c1. We know that x'(0) = 0, which gives us another initial condition.
Substituting x'(0) = 0 into the expression for x':
0 = -g * 0 + c1
0 = c1
Therefore, c1 = 0.
Now we can rewrite the expression for x(t) with the specific values of c1 and c2:
x(t) = -0.5 * g * t^2 + 0 * t + 100
x(t) = -0.5 * g * t^2 + 100
To find how long it takes for the object to hit the ground, we need to find the value of t when x(t) = 0, since the height above the ground will be zero at that time.
Setting x(t) = 0:
-0.5 * g * t^2 + 100 = 0
-0.5 * g * t^2 = -100
g * t^2 = 200
t^2 = 200 / g
t = √(200 / g)
Therefore, it will take t = √(200 / g) units of time for the object to hit the ground.