2. (40 points)A gulf ball (mass-0.05kg) is given an initial speed of 40 m/s at a
ID: 1877333 • Letter: 2
Question
2. (40 points)A gulf ball (mass-0.05kg) is given an initial speed of 40 m/s at an angle of 30° with respect to the horizontal. Assume linear air-resistance: F--bv where b-5x10-3kg/s. (a) (5 points) Write down the x and y components of the Newton's Second Law (b) (5 points) Find the velocity and position as a function of time for the x and y component. (c) (10 points) Find the maximum height reach by the ball and at what time. And find the distance travelled by the ball at this time. (d) (20 points) Find the "approximate" range of the ball and the "approximate" time it landed good to second order in t/t/T,T-m/b Compare the x-distance travelled and time duration during the falling phase to that of rising phase.Explanation / Answer
2. given, m = 0.05 kg
u = 40 m/s
theta = 30 deg wrt horizontal
linear air resistacen, F = -bv
b = 5*10^-3 kg/s
a. from newtons second law
Fx = -b*vx = m*dvx/dt
Fy = -bvy - mg = m*dvy/dt
b. from 1st equation
m*dvx/vx = -b*dt
integerating
m*ln(vx/u*cos(theta)) = -bt
vx = u*cos(theta)*exp(-bt/m)
from 2d equation
-b(vy + mg/b) = m*dvy/dt
dvy/(vy + mg/b) = -b*dt/m
integrating
vy = [(u*sin(theta) + mg/b)exp(-bt/m) - mg/b]
integrating vx
x = mu*cos(theta)*(1 - exp(-bt/m))/b
integrating vy
y = (u*sin(theta) + mg/b)(1 - exp(-bt/m))*m/b - mgt/b
c. for range of ball
y = 0
(u*sin(theta) + mg/b)(1 - exp(-bt/m)) = gt
118.1(1 - exp(-bt/m)) = gt
solving
t = 3.833 s
range = 23.6116071116 m
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