An object is thrown vertically up ward from the ground with initial velocity 1960cm/s Neglecting air resistance, find the maximum height reached and the total time taken to return to the straight point ,/using first order differential equations/

g = -980 cm/s^2

the object reaches its peak at 2 s

average velocity is 980 cm/s
... so peak height is 1960 cm

time up equals time down

Well, well, well... We've got an object going on a wild ride!

To determine the maximum height reached and the total time taken, we can use a little thing called first order differential equations. Hold on tight, because here we go!

Let's start by breaking down the problem. The object is thrown vertically upward, so we know it's going against gravity. The initial velocity is 1960 cm/s, and we need to find the maximum height and total time.

Now, let's dive into the equations. First, we need to know that the acceleration due to gravity is approximately 9.8 m/s² in the downward direction (down, down, down you go!).

Now, we can use the first order differential equation for velocity with respect to time:

dv/dt = -g

Where v is the velocity and g is the acceleration due to gravity (negative because it's going up against gravity).

To solve this differential equation, we integrate both sides:

∫dv = -∫g dt

v = -gt + C

Where C is the constant of integration (oh, those sneaky constants!).

Since the initial velocity is 1960 cm/s (or 19.6 m/s), we can substitute it into our equation and solve for the constant C:

19.6 = -9.8(0) + C
C = 19.6

Now we have our velocity equation:

v = -9.8t + 19.6

To find the maximum height, we know that the velocity at the highest point is 0. So, let's set v = 0 and solve for t:

0 = -9.8t + 19.6
9.8t = 19.6
t = 2 seconds (tick-tock, you reach the highest point in 2 seconds!)

Now, to find the maximum height, we can use the equation:

h = ∫v dt

h = ∫(-9.8t + 19.6) dt
h = -4.9t² + 19.6t + C

Since we're concerned with the maximum height, we substitute t = 2 into the equation:

h = -4.9(2)² + 19.6(2) + C
h = -19.6 + 39.2 + C
h = 19.6 + C

Bingo-bango! The maximum height reached is 19.6 + C (whatever C turns out to be).

But wait, there's more! We need to find the total time taken to return to the starting point. Since we're dealing with a vertically thrown object, it takes the same amount of time to go up as it does to come down (oh, the symmetry of it all!).

So, the total time taken is twice the time it took to reach the highest point:

Total time = 2t
Total time = 2(2)
Total time = 4 seconds

There you have it! The maximum height reached is 19.6 + C (which we'll need more information to find), and the total time taken to return to the starting point is 4 seconds.

Now, excuse me while I join you in catching my breath from all these calculations!

To find the maximum height reached and the total time taken to return to the starting point using first-order differential equations, we can use the equations of motion for vertical motion.

Let's assume that the positive direction is upwards and the negative direction is downwards. The acceleration due to gravity is -9.8 m/s² (negative since it acts downwards).

Step 1: Find the initial velocity in m/s.
Given: Initial velocity = 1960 cm/s
Converting cm/s to m/s: 1960 cm/s = 19.6 m/s

Step 2: Set up the differential equations.
Let's consider y(t) as the displacement of the object at time t.
The first-order differential equations for vertical motion are:
dy/dt = v(t)
dv/dt = a(t)

where:
dy/dt is the derivative of y(t) with respect to time (velocity),
dv/dt is the derivative of v(t) with respect to time (acceleration),
a(t) is the acceleration (constant), which is -9.8 m/s².

Step 3: Solve the differential equations.
Integrating the first equation with respect to t, we get:
∫dy = ∫v(t)dt
y(t) = ∫v(t)dt + C1

Integrating the second equation with respect to t, we get:
∫dv = ∫a(t)dt
v(t) = ∫a(t)dt + C2

Since the acceleration is constant, the integration simplifies to:
v(t) = a(t)t + C2

Step 4: Apply the initial conditions.
At time t = 0, the object is at the ground level, so y(0) = 0.
At time t = 0, the object has an initial velocity of 19.6 m/s, so v(0) = 19.6 m/s.

Using these initial conditions, we can find the values of C1 and C2:
y(0) = 0 = ∫v(0)dt + C1
0 = 0 + C1
C1 = 0

v(0) = 19.6 = a(0) * 0 + C2
19.6 = C2
C2 = 19.6

Step 5: Write the equations using the initial conditions.
The equations become:
y(t) = ∫v(t)dt + 0
y(t) = ∫(a(t)t + 19.6)dt

Integrating, we get:
y(t) = (1/2)a(t)t² + 19.6t

Step 6: Find the maximum height reached.
At the maximum height, the velocity is zero (v = 0). Therefore, we can solve for t when v(t) = 0.

0 = a(t)t + 19.6
at = -19.6
t = -19.6/a

Step 7: Calculate the total time taken to return to the starting point.
The total time taken to return to the starting point is the time it takes for the object to reach the maximum height and then fall back to the ground. This is twice the time it takes to reach the maximum height.

Total time = 2 * t

Substituting the value of t from step 6, we get:
Total time = 2 * (-19.6/a)

Therefore, to find the maximum height reached and the total time taken, you need to know the value of the acceleration, a. Once you have the value of a, substitute it into the equations above to calculate the values.

To solve this problem using first-order differential equations, we need to set up the equation of motion for the object.

Let's assume:
- The upward direction is positive.
- The height of the object above the ground is denoted by 'h.'
- The initial velocity (upward) is given as 1960 cm/s, which we can convert to m/s: v₀ = 1960 cm/s = 19.6 m/s.
- The acceleration due to gravity is g = 9.8 m/s² (approximately constant near the Earth's surface).

Now, we can set up the equation of motion for the object's vertical motion:

Equation 1: dv/dt = -g

This equation represents Newton's second law, which states that the rate of change of velocity with respect to time is equal to the force acting on the object. Here, the negative sign indicates that the acceleration due to gravity is acting in the opposite direction to the initial velocity.

By integrating Equation 1 with respect to time, we can determine the velocity as a function of time:

∫dv = ∫(-g)dt
v(t) = -gt + C

C is the constant of integration. To determine its value, we use the initial condition: at t = 0, v = v₀.

v(0) = -g(0) + C
v₀ = C

Therefore, C = v₀.

The velocity function becomes:
v(t) = v₀ - gt

To find the height function h(t) as a function of time, we integrate the velocity function:

∫dh = ∫(v₀ - gt)dt
h(t) = v₀t - 0.5gt² + D

Here, D is another constant of integration. To determine its value, we use the initial condition: at t = 0, h = 0.

h(0) = v₀(0) - 0.5g(0) + D
0 = 0 + D

Therefore, D = 0.

The height function becomes:
h(t) = v₀t - 0.5gt²

To find the maximum height reached, we know that at the maximum height, the velocity is 0. So, we set v(t) = 0 and solve for t:

0 = v₀ - gt
gt = v₀
t = v₀/g

Substituting this value of t into the height function, we can find the maximum height:

h(max) = v₀t - 0.5gt²
= v₀(v₀/g) - 0.5g(v₀/g)²
= (v₀²)/g - (v₀²)/(2g)
= (v₀²)/(2g)

Now, let's calculate the maximum height reached. Substituting the given values, we have:
v₀ = 19.6 m/s, g = 9.8 m/s².

h(max) = (19.6²)/(2 * 9.8)
= 196/2
= 98 meters

Therefore, the maximum height reached is 98 meters.

To find the total time taken to return to the starting point, we can use the fact that the object reaches the same height at the start and end of its trajectory.

Therefore, h(t) = h(0) = 0.

Using the height function, we have:
v₀t - 0.5gt² = 0
t(v₀ - 0.5gt) = 0
t = 0 or v₀ - 0.5gt = 0

Since t = 0 corresponds to the initial time, we are interested in the second solution:

v₀ - 0.5gt = 0
v₀ = 0.5gt
t = v₀ / (0.5g)

Substituting the given values, we have:
v₀ = 19.6 m/s, g = 9.8 m/s².

t = 19.6 / (0.5 * 9.8)
= 19.6 / 4.9
= 4 seconds

Therefore, the total time taken to return to the starting point is 4 seconds.