Newton's equation for a free falling object with air resistance of mass m kilograms says that its velocity v(t) satisfies the DE

mv'(t) = mg - kv(t)

where g = 9.8 \frac{m}{s^2} and v(t) is measured in \frac{m}{s}. Suppose that k=2 \frac{kg}{sec} and that a ball of mass m = 1 kg falls with initial velocity v(0)=0. Round to the nearest tenth.

(a) What is its velocity after 1.5 seconds?
So far I have this, but not sure where to go from this:

dv/dt + 2v = 9.8

(b) What is the terminal velocity?

Your DE yields

v = c e^(-2t) + 4.9
v(0) = 0 means c = -4.9, so

v(t) = 4.9 - 4.9e^(-2t)

(a) plug in t=1.5
(b) v(∞) = 4.9

To find the velocity of the object after 1.5 seconds, we need to solve the differential equation:

mv'(t) = mg - kv(t)

First, let's rewrite the equation in terms of acceleration (a) instead of velocity (v):

ma = mg - kv(t)

Since a = v'(t), we can rearrange the equation as:

mv'(t) + kv(t) = mg

Now, substitute the given values of m = 1 kg, k = 2 kg/sec, and g = 9.8 m/s² into the equation:

v'(t) + 2v(t) = 9.8

This is a first-order linear ordinary differential equation (ODE) in the form of:

v'(t) + p(t)v(t) = q(t)

where p(t) = 2 and q(t) = 9.8.

To solve this type of equation, we can use an integrating factor. The integrating factor (IF) is given by:

IF = e^(∫p(t)dt) = e^(∫2dt) = e^(2t)

Now, multiply both sides of the equation by the integrating factor:

e^(2t)v'(t) + 2e^(2t)v(t) = 9.8e^(2t)

Apply the product rule on the left side to simplify:

(d/dt)(e^(2t)v(t)) = 9.8e^(2t)

Integrate both sides with respect to t:

∫(d/dt)(e^(2t)v(t)) dt = ∫9.8e^(2t) dt

e^(2t)v(t) = ∫9.8e^(2t) dt

Integrating the right side of the equation gives:

e^(2t)v(t) = 4.9e^(2t) + C

To find the constant of integration (C), we need to use the initial condition v(0) = 0:

e^(2(0))v(0) = 4.9e^(2(0)) + C
v(0) = 0 = 4.9 + C
C = -4.9

Substitute the value of C back into the equation:

e^(2t)v(t) = 4.9e^(2t) - 4.9

Finally, isolate v(t):

v(t) = (4.9e^(2t) - 4.9) / e^(2t)

Now, we can substitute t = 1.5 seconds into this equation to find the velocity after 1.5 seconds:

v(1.5) = (4.9e^(2(1.5)) - 4.9) / e^(2(1.5))

Evaluate this expression using a calculator or software to get the velocity after 1.5 seconds.

For part (b), the terminal velocity occurs when the velocity (v) becomes constant, meaning v'(t) = 0. Rearranging the original differential equation, we have:

ma = mg - kv(t)
0 = mg - kv(t)

Solving for v(t), we get:

v(t) = mg / k

Substituting the given values of m = 1 kg, g = 9.8 m/s², and k = 2 kg/sec into the equation:

v(t) = (1 kg)(9.8 m/s²) / (2 kg/sec)

Evaluate this expression to find the terminal velocity.