Suppose you throw an object from a great height, so that it reaches very nearly terminal velocity by time it hits the ground. By measuring the impact, you determine that this terminal velocity is -49 mi sec.
A. Write the equation representing the velocity v(t) of the object at time t seconds given the initial velocity v0 and the fact that acceleration due to gravity 9.8 m/sec2. (Here, assume you're modeling the falling body with the differential equation dy/dt = g-kv, and use the resulting formula or v(t) found in the Tutorial. Of course, you can derive it if you'd like.)
B. Determine the value of k, the "continuous percentage growth rate" from the velocity equation, by utilizing the information given concerning the terminal velocity.
C. Using the value of k you derived above, at what velocity must the object be thrown upward if you want it to reach its peak height after 3 sec? Approximate your solution to three decimal places, and justify your answer.
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A. To represent the velocity v(t) of the object at time t seconds, we can use the differential equation dy/dt = g - kv, where g is the acceleration due to gravity (9.8 m/s^2) and v is the velocity of the object.
Taking the derivative with respect to t of v, we get dv/dt = -k * v.
Rearranging the equation, we have dv/v = -k * dt.
Integrating both sides, we get ∫dv/v = -k * ∫dt.
The integral of dv/v is ln|v|, and the integral of dt is t. Therefore, the equation becomes ln|v| = -k * t + C, where C is the constant of integration.
Exponentiating both sides, we get |v| = e^(-kt+C).
Since we are interested in the velocity, which cannot be negative, |v| becomes v. So, we have v = e^(-kt+C).
To eliminate the constant C, we can use the fact that the object reaches its terminal velocity of -49 m/s. When t approaches infinity, the exponential term e^(-kt) becomes very close to 0, and v approaches -49 m/s.
Thus, we have -49 = e^(0+C).
Simplifying, we get -49 = e^C.
Taking the natural logarithm of both sides, we have ln(-49) = C.
Therefore, the equation representing the velocity of the object at time t seconds is v(t) = e^(-kt+ln(-49)).
B. We know that the terminal velocity is -49 m/s. Plugging in the values into the equation, we have -49 = e^(-k * ∞ + ln(-49)).
Since e^(-k * ∞) is approaching 0, we can ignore it. Therefore, ln(-49) = -k * ∞.
Taking the limit as t approaches infinity, we get k = lim(t→∞) (ln(-49) / -t).
C. To find the velocity at which the object must be thrown upward to reach its peak height after 3 seconds, we can use the equation v(t) = e^(-kt+ln(-49)).
Since we want the object to reach its peak height after 3 seconds, we can set t = 3 and solve for v.
Therefore, v(3) = e^(-k*3+ln(-49)).
Approximating the value of k from part B, we substitute it into the equation v(3) = e^(-k*3+ln(-49)) and solve for v(3).
A. To represent the velocity v(t) of the object at time t seconds, we will use the following differential equation:
dy/dt = g - kv
Where:
- dy/dt represents the rate of change of the displacement y with respect to time.
- g is the acceleration due to gravity (which is negative in this case as it acts downward, so g = -9.8 m/sec^2).
- k is a constant that represents the air resistance.
To convert the units of gravitational acceleration from m/sec^2 to mi/sec^2, we use the conversion factor: 1 m/sec^2 ≈ 0.0372823 mi/sec^2.
So, the equation becomes:
dy/dt = (-9.8) * 0.0372823 - kv
B. We are given that the terminal velocity of the object is -49 mi/sec. At terminal velocity, the acceleration due to gravity and air resistance balance out, meaning that the net force acting on the object is zero. Since force is related to acceleration by Newton's second law (F = ma), we have:
0 = mg - kv
Solving for k:
k = mg / v
Using the given acceleration due to gravity in mi/sec^2 and the terminal velocity, we have:
k = (-9.8) * 0.0372823 / (-49)
C. To determine the initial upward velocity needed for the object to reach its peak height after 3 seconds, we can use the formula for velocity:
v(t) = v0 * e^(kt)
Where:
- v(t) is the velocity at time t,
- v0 is the initial velocity,
- e is the base of the natural logarithm.
Considering the upward motion, the initial velocity would be positive. Rearranging the equation to solve for v0:
v0 = v(t) / e^(kt)
Plugging in the values:
- t = 3 seconds,
- k = the value derived in part B,
- v(t) = 0 (at the peak height, the velocity is 0),
We get:
v0 = 0 / e^(k*3)
Simplifying further:
v0 = 0
Therefore, the object must be thrown upward with an initial velocity of 0 mi/sec to reach its peak height after 3 seconds.
v = dy/dt
so, if v = g-kv
this is just a linear DEm with solution
v = ce^(-3t) + g/k
since v(0) = v0,
c = v0 - g/k
That may not look exactly like the formula you derived, but it is the same, up to assigning the constants.
for large t, e^(-3t) = 0, so we have
g/k = -49
Now you can answer C.