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.
this problem is discussed in detail at
http://tutorial.math.lamar.edu/Classes/DE/Modeling.aspx
just scroll down to where it talks about falling bodies.
To solve these problems, we can use Newton's second law of motion, which states that the sum of all forces acting on an object is equal to the product of its mass and acceleration.
Let's break this down step by step:
a) Finding the velocity 0.1 seconds after the object is thrown:
First, we need to calculate the net force acting on the object at that time. The net force is the force acting on the object minus the force of air resistance.
The force acting on the object is equal to its weight, which can be calculated using the formula W = mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 980 cm/sec^2).
W = (mass) x (acceleration due to gravity)
W = 100g x 980 cm/sec^2
W = 98000 g cm/sec^2
Since we are given that the force of air resistance is numerically equal to 200v (in dynes) and the object is at rest (v = 0) initially, the force of air resistance is also zero at this point.
Therefore, the net force is equal to the force acting on the object:
Net Force = 98000 g cm/sec^2
Next, we can use Newton's second law to find the acceleration of the object:
Net Force = mass x acceleration
98000 g cm/sec^2 = 100g x acceleration
Dividing both sides by 100g:
980 g cm/sec^2 = acceleration
Now, we can find the velocity 0.1 seconds after the object is thrown. We'll use the formula for average acceleration:
v = u + at
where v is the final velocity, u is the initial velocity (150 cm/sec), a is the acceleration, and t is the time (0.1 seconds).
Using the values we have:
v = 150 cm/sec + (980 cm/sec^2) x (0.1 sec)
v = 150 cm/sec + 98 cm/sec
v = 248 cm/sec
Therefore, the velocity of the object 0.1 seconds after it is thrown vertically upward is 248 cm/sec.
b) Finding the velocity 0.1 seconds after the object stops rising and starts falling:
When the object reaches its maximum height and begins to fall, the force of air resistance is no longer zero. Using the given information that the force of air resistance is numerically equal to 200v (in dynes), we can calculate the force of air resistance when the object starts falling.
At this point, the velocity of the object is also given as 150 cm/sec in the upward direction.
Force of air resistance = 200v
Force of air resistance = 200 x 150 cm/sec
Force of air resistance = 30000 cm dynes
To find the net force, we'll subtract the force of air resistance from the force acting on the object:
Net Force = 98000 g cm/sec^2 - 30000 cm dynes
Converting the force of air resistance to cm/sec^2 (1 dyne = 1 g cm/sec^2):
Net Force = 98000 g cm/sec^2 - 30000 g cm/sec^2
Net Force = 68000 g cm/sec^2
Applying Newton's second law:
Net Force = mass x acceleration
68000 g cm/sec^2 = 100g x acceleration
Dividing both sides by 100g:
680 g cm/sec^2 = acceleration
Using the same formula as before:
v = u + at
v = 150 cm/sec + (680 cm/sec^2) x (0.1 sec)
v = 150 cm/sec + 68 cm/sec
v = 218 cm/sec
Therefore, the velocity of the object 0.1 seconds after it stops rising and starts falling is 218 cm/sec.