somebody please help me find the solution totally lost....
An object of mass 100g is thrown vertically upward from a point 60cm above the earths surface with an initial velocity of 150cm/sec. It rises briefly and then falls vertically to the earth, all of which time it is acted on by air resistance that is numerically equal to 200v (in dynes), where v is the velocity (in cm/sec).
a) find the velocity 0.1sec after the object is thrown.
b)find the velocity 0.1sec after the object stops rising and start falling.
Oh, don't worry! I'm here to help you find the solution, and hopefully bring a smile to your face along the way! Let's solve this step by step:
a) To find the velocity 0.1 second after the object is thrown, we need to consider the initial velocity and the acceleration caused by air resistance. Now, I know you mentioned the force of air resistance being 200v (in dynes), but to keep things simple, let's convert it to grams instead. So, 200v (in dynes) would be 200v (in grams/cm/sec^2).
Now, the force due to air resistance can also be written as F = ma (force equals mass times acceleration). Since the mass of the object is 100g, and the acceleration is 200v (in grams/cm/sec^2), we can rewrite F = ma as 200v = 100a.
Since we have the formula for acceleration, we can use it to find the velocity. The equation for velocity is v = u + at, where "u" is the initial velocity, "a" is the acceleration, and "t" is the time.
In this case, the initial velocity (u) is given as 150 cm/sec, and the acceleration (a) is obtained from 200v = 100a. So, let's solve for a first:
200v = 100a
Divide both sides by 100:
2v = a
Now, we can use this value of acceleration to find the velocity after 0.1 second:
v = u + at
v = 150 + 2v x 0.1
v = 150 + 0.2v
Subtract 0.2v from both sides:
0.8v = 150
Divide both sides by 0.8:
v = 187.5 cm/sec
So, the velocity 0.1 second after the object is thrown is 187.5 cm/sec.
b) To find the velocity 0.1 second after the object stops rising and starts falling, we need to consider the gravitational force acting on the object. In this case, the acceleration due to gravity is approximately 980 cm/sec^2.
Since the object has stopped rising, its velocity is momentarily zero. So, we can use the formula v = u + at again:
0 = 0 + 980 x t
t = 0.1 second
So, the time it takes for the object to stop rising is also 0.1 second.
After the object starts falling, the air resistance would be acting in the opposite direction of the velocity. We can write this as -200v (in grams/cm/sec^2). Assuming the object is falling downward, the acceleration caused by air resistance can be expressed as -200v = 100a.
Using the formula again to find the velocity after 0.1 second:
v = u + at
v = 0 + (-2v) x 0.1
v = -0.2v
Add 0.2v to both sides:
1.2v = 0
Divide both sides by 1.2:
v = 0
So, the velocity 0.1 second after the object stops rising and starts falling is 0 cm/sec.
I hope that helped! If you have any more questions, feel free to ask!
To find the velocity of the object at different points in time, we can use the equations of motion.
a) Find the velocity 0.1sec after the object is thrown:
Given:
Initial velocity, u = 150 cm/sec (positive value because it is thrown upward)
Acceleration, a = -200v (due to air resistance)
Time, t = 0.1 sec
Using the equation: v = u + at
Substituting the given values:
v = 150 + (-200v) * 0.1
Simplifying the equation:
v = 150 - 20v
Combine the terms with 'v' on one side and constant terms on the other side:
v + 20v = 150
21v = 150
Divide both sides by 21:
v = 150 / 21
v ≈ 7.14 cm/sec
Therefore, the velocity of the object 0.1 seconds after it was thrown is approximately 7.14 cm/sec.
b) Find the velocity 0.1 sec after the object stops rising and starts falling:
When the object stops rising and starts falling, its velocity at that point is zero. So we need to find the time it takes for the object to reach that point.
Using the equation for displacement: s = ut + 0.5at^2
Given:
Initial velocity, u = 150 cm/sec
Acceleration, a = -200v (due to air resistance)
t = unknown
At the point where the object stops rising and starts falling, the displacement (s) is 60 cm (the height from which it was thrown).
60 = 150t + 0.5(-200v)t^2
Since we are looking for the time it takes for the object to reach this point (when velocity is zero), we can set v = 0.
60 = 150t + 0.5(-200)(0)t^2
60 = 150t
Divide both sides by 150:
t = 60/150
t = 0.4 sec
Now that we have the time when the object stops rising and starts falling, we can find its velocity 0.1 seconds after that.
Using the equation: v = u + at
Substituting the given values:
v = 150 + (-200v) * (0.1 + 0.4)
Simplifying the equation:
v = 150 - 200v * 0.5
Combine the terms with 'v' on one side and constant terms on the other side:
v + 200v * 0.5 = 150
v * (1 + 0.5*200) = 150
v * (1 + 100) = 150
v * 101 = 150
Divide both sides by 101:
v = 150 / 101
v ≈ 1.48 cm/sec
Therefore, the velocity of the object 0.1 seconds after it stops rising and starts falling is approximately 1.48 cm/sec.
To find the velocity of the object at different times, we can use the equation of motion for a vertically thrown object under the influence of air resistance:
m(dv/dt) = mg - kv,
Where:
m = mass of the object (100g = 0.1kg)
g = acceleration due to gravity (9.8m/s²)
k = constant for air resistance (k = 200)
a) To find the velocity 0.1 seconds after the object is thrown, we need to integrate the equation of motion from t = 0 to t = 0.1 seconds.
By integrating the equation, we get:
m∫[(dv)/(mg - kv)] = ∫dt
Simplifying and solving the integral, we have:
(1/k)ln((mg - kv)/mg) = t + C,
Where C is the constant of integration.
Rearranging the equation and substituting the given values, we can solve for C:
ln((mg - kv)/mg) = kt + C
(mg - kv)/mg = e^(kt + C)
Now, substitute the initial conditions: t = 0, v = 150 cm/s.
(0.1*9.8 - 200v)/0.1*9.8 = e^(0 + C)
(0.98 - 200*150)/0.98 = e^C
We can solve for C to get:
C = ln((0.98 - 200*150)/0.98)
Now, substitute the values of t = 0.1 seconds into the equation and solve for v:
(0.1*9.8 - 200v)/0.1*9.8 = e^(0.1k + C)
Rearrange the equation to solve for v:
v = (0.1*9.8 - 0.1*9.8*e^(0.1k + C))/200
Substitute the values of k and C to get the final answer.
b) To find the velocity 0.1 seconds after the object stops rising and starts falling, we can use the same equation of motion. However, this time the initial velocity is zero because the object comes to a stop before starting to fall.
Follow the same steps as in part (a), but this time substitute the initial velocity (v = 0 cm/s) and solve for the velocity at t = 0.1 seconds.