Question

A ball of mass m is thrown at an angle θ to the ground with a coefficient of friction μ. There is friction acting on the ball during the partially elastic collision. The coefficient of restitution for the impact is e.

At what angle θ is the range d of the ball maximum?
Data: e = 0.25. μ = 0.4.

Answer 1

Tan 2θ = 1/[μ* (1+e)] = 2
. SOLUTION.
mass of the ball = m.
Velocity after bouncing = v.
Angle of bouncing = θ2.
Velocity before bouncing = u.
time duration of contact = t.
Range d = 2 v^2 sin θ2 cosθ2/ g. ----(1)
v sin θ2= e * u sin θ. ----(2)
N * t = m(v sinθ2 + u sinθ) ---(3)
m(u cosθ - v cos θ2) = f * t ---(4)
f = μ N ----(5)

Solving these we get:
d = (2u^2/g)*e sin θ*[cosθ - μ(1+e)*sinθ ]
d = e u^2/g* [sin 2θ + μ(1+e)(cos 2θ - 1)]
Differentiate d wrt θ to find max d.
Tan 2θ = 1/[μ(1+e)].

For the data given tan 2θ = 2.