A ballast is dropped from a stationary hotair balloon that is hovering at an altitude of 500 ft. The velocity of the ballast after t sec is -32t ft/sec. a) Find the height h(t) of the ballast from the ground at time t. Hint: h'(t) = -32t and h(0) = 500. h(t) = 500 - 16t^2 ft.
b) When will the ballast strike the ground? __________sec
c) Find the velocity of the ballast when it hits the ground. ____________ft/sec.
(a) is correct
(b) solve for t when h=0
(c) use the above t to get v
a) To find the height function h(t) of the ballast, we can integrate the velocity function -32t ft/sec with respect to time.
∫(-32t) dt = -16t^2 + C
Since h(0) = 500, we can use this initial condition to solve for C:
500 = -16(0)^2 + C
500 = C
Therefore, the height function of the ballast is:
h(t) = -16t^2 + 500 ft
b) To determine when the ballast will strike the ground, we set h(t) = 0 and solve for t:
-16t^2 + 500 = 0
16t^2 = 500
t^2 = 500/16
t^2 = 31.25
t = √(31.25)
t ≈ 5.59
Hence, the ballast will strike the ground after approximately 5.59 seconds.
c) The velocity of the ballast when it hits the ground can be found by evaluating the velocity function at t = 5.59:
v(t) = -32t ft/sec
v(5.59) = -32(5.59) ft/sec
v ≈ -178.88 ft/sec
Therefore, the velocity of the ballast when it hits the ground is approximately -178.88 ft/sec.
To answer these questions, we will use the given information and apply principles of calculus. Let's break it down step by step:
a) Given that the velocity of the ballast after t seconds is -32t ft/sec, we need to find the height h(t) of the ballast from the ground at time t. We can start by integrating the given velocity function to find the position function.
The integral of -32t dt gives us -16t^2 + C, where C is the constant of integration. Since we are given that h(0) = 500, we can substitute this condition into the position equation to solve for C:
500 = -16(0)^2 + C
500 = 0 + C
C = 500
Now we have the position function:
h(t) = -16t^2 + 500 ft
So, the height of the ballast from the ground at any time t is h(t) = 500 - 16t^2 ft.
b) To find when the ballast will strike the ground, we need to find the time when h(t) = 0 (since the height from the ground is 0 when the ballast strikes the ground). We can set up the equation:
0 = 500 - 16t^2
Simplifying, we have:
16t^2 = 500
Dividing both sides by 16 gives us:
t^2 = 31.25
Taking the square root of both sides:
t = ±√31.25
Since time cannot be negative in this context, we only consider the positive square root:
t ≈ 5.59 seconds
So, the ballast will strike the ground at approximately 5.59 seconds.
c) To find the velocity of the ballast when it hits the ground, we can take the derivative of the position function h(t) with respect to time:
h'(t) = -32t ft/sec
Substituting t = 5.59 into h'(t), we get:
h'(5.59) = -32(5.59) ft/sec
h'(5.59) ≈ -178.88 ft/sec
Therefore, the velocity of the ballast when it hits the ground is approximately -178.88 ft/sec. Note that the negative sign indicates that the ballast is moving downward.