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 m/sec.
A. Write the equation representing the velocity v(t) of the object at time t seconds given the initial velocity v sub 0 and the fact that acceleration due to gravity is -9.8 m/sec^2.
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?
A. The equation representing the velocity v(t) of the object at time t seconds can be derived using the principles of physics. Since the object is falling under the influence of gravity, we can use the equation of motion:
v(t) = v0 + a * t
Where:
v(t) is the velocity of the object at time t
v0 is the initial velocity of the object
a is the acceleration due to gravity (-9.8 m/sec^2)
Therefore, the equation representing the velocity v(t) of the object at time t seconds is:
v(t) = v0 - 9.8t
B. To determine the value of k, the "continuous percentage growth rate" from the velocity equation, we can use the fact that at terminal velocity, the acceleration is zero. Since the object is falling, the acceleration is in the opposite direction to the initial velocity. The terminal velocity, denoted as Vt, is reached when the acceleration due to gravity (-9.8 m/sec^2) exactly cancels out the initial velocity v0.
So, at terminal velocity, we have:
v(t) = Vt = v0 - 9.8t
Since the object is falling, v0 is positive, and therefore, Vt is negative. Given that the terminal velocity is -49 m/sec, we can set Vt = -49 and solve for v0:
-49 = v0 - 9.8t
C. Using the value of k derived above, we can determine the velocity at which the object must be thrown upward to reach its peak height after 3 seconds. The peak height is reached when the velocity becomes zero. Therefore, we need to find the initial velocity v0 that will result in a velocity of zero after 3 seconds.
Using the equation v(t) = v0 - 9.8t and setting t = 3, we have:
0 = v0 - 9.8 * 3
Solving for v0:
v0 = 9.8 * 3
v0 = 29.4 m/sec
Therefore, the object must be thrown upward with an initial velocity of 29.4 m/sec to reach its peak height after 3 seconds.
A. The equation representing the velocity of the object at time t seconds can be derived using the equation of motion:
v(t) = v₀ + a*t
where:
v(t) is the velocity at time t
v₀ is the initial velocity
a is the acceleration
In this case, the initial velocity v₀ is not given explicitly, but it can be assumed to be 0 since the object starts from rest. The acceleration due to gravity, a, is -9.8 m/sec² (negative because it acts in the opposite direction of positive velocity).
Thus, the equation representing the velocity v(t) is:
v(t) = 0 + (-9.8)*t
v(t) = -9.8*t
B. The terminal velocity, -49 m/sec, is the maximum velocity the object reaches when it stops accelerating under the influence of gravity and experiences an equal and opposite force due to air resistance. At terminal velocity, the net force on the object is zero.
To determine the value of k, which represents the continuous percentage growth rate, we can use the exponential decay formula:
v(t) = v₀ * e^(k*t)
At terminal velocity, the object no longer accelerates, so the value of v(t) remains constant. Therefore, we can write:
v(t) = -49 * e^(k*t)
Since v(t) remains constant, its derivative with respect to t is 0. Taking the derivative:
v'(t) = 0 = -49 * k * e^(k*t)
Since e^(k*t) is never zero, we can simplify the equation to:
0 = -49 * k
Therefore, k = 0.
C. If we want the object to reach its peak height after 3 seconds, we need to determine the initial velocity v₀. At the peak height, the velocity is 0, and the object undergoes free fall for 3 seconds.
Using the equation from part A:
v(t) = v₀ - 9.8*t
We can substitute t = 3 and v(t) = 0:
0 = v₀ - 9.8*3
Solving for v₀:
v₀ = 9.8*3
v₀ = 29.4 m/sec
Therefore, the object must be thrown upward with an initial velocity of 29.4 m/sec to reach its peak height after 3 seconds.
v(t) = v0 - 9.8t
see what you can do with that...