A particle of mass 100 gm is thrown vertically upwards with a velocity of 7 m/s. Air resistance is also acting on the particle according to the equation F = -0.2v^2. ( F = Force, v = velocity of the particle ). Find how much high the ball rises.
Please answer it.
Thanks in advance.
I'm having a bit of trouble with F. Force is in units of N = kg-m/s^2
Having v^2 there makes the units m^2/s^2. How do you get rid of the extra meters?
Anyway, when you fix that, just remember that F = ma, so a = F/m. Just add that to the -9.8 m/s^2 of gravity, and find the vertex of the resulting parabola.
This is a bit tricky.
air resistance is indeed dependent on velocity squared.
initial KE=final height- friction work
1/2 m 7^2=mgh - friction
friction losses:
during falling, F=ma=m(dv/dt)
but we know F=mg-kv^2 or
m(dv/dt)=mg-kv^2
so now you have a differential equation in v.
v'=g-k/m v^2
This is not a trivial problems. See page 10 at https://prettygoodphysics.wikispaces.com/file/view/DifferentialEquations.pdf
To find the height that the ball rises, we need to use the equations of motion. The key is to determine the time it takes for the ball to reach its highest point.
First, let's find the acceleration of the ball due to air resistance. We have the force equation as F = -0.2v^2, where F is the force and v is the velocity of the particle. However, the force is related to the acceleration by Newton's second law (F = ma), so we can rewrite the equation as ma = -0.2v^2, where m is the mass of the particle.
Next, let's calculate the acceleration. We know that acceleration is the derivative of velocity with respect to time, so we can rewrite the equation as m(dv/dt) = -0.2v^2.
Since we are looking for the height, we need to integrate this equation with respect to time. By applying separation of variables, we have:
(1/m) * dv/v^2 = -0.2dt.
Integrating both sides:
∫ (1/m) * dv/v^2 = ∫ -0.2 dt.
This gives us:
(-1/mv) = -0.2t + C,
where C is the constant of integration.
Now, we can solve for t, which represents the time taken to reach the highest point. When the ball reaches the highest point, its velocity is 0. Substituting into the equation:
(-1/m(0)) = -0.2t + C.
Since m is positive, we have:
0 = -0.2t + C.
Solving for t, we find:
t = C/0.2.
To determine the value of C, we need initial conditions. The particle is thrown vertically upwards with a velocity of 7 m/s. We can choose this as an initial condition, at t = 0 and v = 7. Substituting these values into our equation:
0 = -0.2(0) + C,
C = 0.
Therefore, C = 0, and our equation becomes:
t = 0/0.2,
t = 0.
This tells us that it takes 0 seconds for the ball to reach its highest point.
Now that we know the time taken, we need to find the height the ball rises. We can use the equation of motion:
h = ut + (1/2)gt^2,
where h represents height, u represents initial velocity, g represents acceleration due to gravity, and t represents time.
Since the particle is thrown vertically upwards, the acceleration due to gravity should be negative (-9.8 m/s^2).
Plugging in the values, we have:
h = (7)(0) + (1/2)(-9.8)(0)^2,
h = 0.
Therefore, the ball does not rise to any height due to the air resistance acting against the upward motion.