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?
x"=-9.8
x'=INT (-9.8)=-9.8t+x'(0)
x= INT (-9.8t+x'(0))
= x'(0)*t-1/2 9.8t^2 + x(0)
b) given x(0)=100, x'(0)=0
x=-4.9t^2 if x=-100
then t= sqrt(100/4.9) when it hits the ground.
a) To find a general expression for the height of the object above the ground with respect to time, t, we need to solve the differential equation x" = -g.
We can start by integrating both sides of the equation with respect to time:
∫x" dt = ∫-g dt
Integrating x" with respect to time gives us x' (the velocity):
x' = -gt + C1
Where C1 is the constant of integration.
Next, we integrate x' with respect to time to find x (the height):
x = -0.5gt^2 + C1t + C2
Where C2 is another constant of integration.
Now, we need to determine the values of C1 and C2 by applying the initial conditions.
b) If the object is dropped from a height of 100 units, we know that x(0) = 100 and x'(0) = 0.
Substituting t = 0 into the equation for x, we get:
x(0) = -0.5g(0)^2 + C1(0) + C2
100 = C2
Using the initial condition x'(0) = 0, we can solve for C1:
x' = -gt + C1
0 = -g(0) + C1
C1 = 0
Therefore, our general expression for x(t) becomes:
x(t) = -0.5gt^2 + 100
To find how long it will take for the object to hit the ground, we set x(t) equal to 0 and solve for t:
0 = -0.5gt^2 + 100
Simplifying the equation, we get:
0.5gt^2 = 100
gt^2 = 200
t^2 = 200 / g
t = √(200 / g)
So, it will take approximately √(200 / g) units of time for the object to hit the ground.