A ball of mass m is thrown vertically upwards with initial speed u, and travels upwards under the influence of gravity and air resistance. Use the quadratic model of air resistance with the ball modelled as a sphere of effective diameter D.
Q1) Apply Newton’s second law to obtain the equationdv/dt = −k(v^2 + g/k)where v is the speed of the ball, t is the time elapsed since the ball was released, g is the magnitude of the acceleration due to gravity, andk = c_2D_2/m is a constant.
Q2)Solve the differential equation and apply the initial condition to find thetime t in terms of the speed v and the constants given above.
Q3) Use your equation to show that the time taken to reach the maximum height attained by the ball, tmax, is given by tmax = 1/√(gk) arctan (√(k/g)u)
Q1) To apply Newton's second law to the motion of the ball, we need to consider the forces acting on it. In this case, the two main forces are gravity and air resistance.
The force of gravity acting on the ball can be represented by the equation F_gravity = -mg, where m is the mass of the ball and g is the acceleration due to gravity.
The force of air resistance can be modeled using the quadratic model, which states that the resistance force is proportional to the square of the speed of the object. In this case, we can represent air resistance as F_air = -kv^2, where k is a constant.
Applying Newton's second law (F_net = ma) to the vertical motion of the ball, we have:
ma = F_gravity + F_air
m(dv/dt) = -mg - kv^2
Rearranging the equation and factoring out common factors, we get:
dv/dt = (-mg - kv^2) / m
dv/dt = -g - (k/m) * v^2
Since the expression (-g - (k/m) * v^2) is a constant, we can simplify it further by defining another constant, k = c_2D_2/m. Thus, we arrive at the equation:
dv/dt = -g - k * v^2
or
dv/dt = -k(v^2 + g/k)
Q2) To solve this differential equation, we need to separate variables and integrate both sides. Let's denote the integral of dv/dt with respect to t as V and integrate the equation:
∫(1 / (v^2 + g/k)) dv = ∫-k dt
To integrate the left-hand side, we use the substitution u = v^2 + g/k:
∫(1 / (u)) dv = ∫-k dt
Using the formula for the integral of 1/u, we have:
ln|u| = -kt + C
Substituting the value of u back, we get:
ln|v^2 + g/k| = -kt + C
Applying the initial condition, let's assume the ball was released at time t = 0 with an initial velocity u:
At t = 0, v = u, so substituting these values into the equation, we have:
ln|(u^2 + g/k)| = C
Taking the exponential of both sides, we get:
u^2 + g/k = e^(C)
Since e^(C) is just another constant, let's denote it as A:
u^2 + g/k = A
Solving for v, we have:
v^2 = A - g/k
Taking the square root of both sides, we get:
v = ±√(A - g/k)
Q3) To find the time taken to reach the maximum height attained by the ball, we need to find the time when the velocity becomes zero (v = 0). Substituting v = 0 in the equation derived above:
0 = ±√(A - g/k)
Squaring both sides, we get:
0 = A - g/k
Solving for A, we have:
A = g/k
Substituting this value of A into the equation for v, we get:
v = ±√(g/k - g/k)
Simplifying further:
v = ±√(0)
Thus, at maximum height, the velocity is 0. Now, we can substitute the initial condition v = u into the equation to find the time taken to reach maximum height:
u = ±√(0 - g/k)
u = ±√(- g/k)
Since u is the initial upward velocity, it is positive. Therefore, we consider only the positive square root. Simplifying further, we get:
u = √(g/k)
Taking the inverse tangent of both sides, we obtain:
arctan(u/√(g/k)) = arctan(1)
arctan(u/√(g/k)) = π/4
Therefore, the time taken to reach the maximum height is:
t_max = 1/√(gk) arctan(√(k/g)u)
Please note that the above derivation assumes certain assumptions and approximations, as stated in the question, such as the quadratic model of air resistance and neglecting other factors that may affect the motion of the ball.