A stone is thrown straight up 6 feet from the ground (it is released 18 feet above the ground). When it is released it is travelling at the rate of 100 feet per second.
a) Find the velocity function.
b) Find the position function.
c) How high does the stone go?
d) How long does it take for the stone to reach the ground?
What does "thrown up 6 feet from the ground(it is released 18 feet above the ground)" mean.
To solve these questions, we will use the basic principles of calculus, specifically derivatives and integrals. Let's tackle each question one by one:
a) Find the velocity function:
Velocity is the derivative of the position function with respect to time. Since the stone is thrown straight up, the only relevant variable is time, denoted as "t." The initial velocity of the stone is given as 100 feet per second when it is released. Thus, the velocity function is:
v(t) = -32t + 100
Here, the "-32t" term comes from the acceleration due to gravity (32 feet per second squared) acting in the opposite direction as the stone moves up.
b) Find the position function:
The position function can be found by integrating the velocity function. Since the velocity is an expression with respect to time, integrating it will give us the position function. We integrate the velocity function to find the displacement function, and then add the initial position of the stone (18 feet above the ground) to get the position function:
s(t) = ∫ (v(t)) dt + 18
= ∫ (-32t + 100) dt + 18
= -16t^2 + 100t + 18
Therefore, the position function is:
s(t) = -16t^2 + 100t + 18
c) How high does the stone go?
To find the maximum height reached by the stone, we need to find the vertex of the parabolic position function. We can do this by finding the value of "t" that corresponds to the vertex. The vertex occurs at the maximum point of the parabola, where the derivative of the position function is equal to zero:
s'(t) = 0
Differentiating the position function, we get:
s'(t) = -32t + 100
Setting this equal to zero, we solve for "t":
-32t + 100 = 0
-32t = -100
t = 3.125 seconds
Substituting this value into the position function:
s(3.125) = -16(3.125)^2 + 100(3.125) + 18
= -156.25 + 312.5 + 18
= 174.25 feet
Hence, the stone reaches a maximum height of 174.25 feet.
d) How long does it take for the stone to reach the ground?
To determine when the stone hits the ground, we need to find the value of "t" when the position function equals zero:
s(t) = -16t^2 + 100t + 18 = 0
Solving this quadratic equation will give us the two roots, which represent the times when the stone is at ground level. We can use the quadratic formula to find these values:
t = (-b ± √(b^2 - 4ac)) / 2a
Here, a = -16, b = 100, and c = 18. Plugging these values into the quadratic formula:
t = (-100 ± √(100^2 - 4(-16)(18))) / 2(-16)
t = (-100 ± √(10000 + 1152)) / -32
t = (-100 ± √11152) / -32
After calculating these values, we find:
t ≈ -0.309 and t ≈ 6.809
Since time cannot be negative, we conclude that the stone reaches the ground after approximately 6.809 seconds.
To summarize:
a) The velocity function is v(t) = -32t + 100.
b) The position function is s(t) = -16t^2 + 100t + 18.
c) The stone reaches a maximum height of approximately 174.25 feet.
d) The stone takes approximately 6.809 seconds to reach the ground.