Daniel B. answered 12/26/20
A retired computer professional to teach math, physics
I am assuming that this question came from chegg.com, and I will use Figure 5
on that website.
Let W(θ) be the work performed by gravity by the time the snowball reaches n arbitrary angle θ.
Let dW(θ) be its increment due to the angle increasing by dθ.
As usual, dθ is small enough that to first order we can assume the snowball
to loose no mass while traversing the angle dθ.
dW(θ) = m(θ) g R dθ sin(θ)
where
m(θ) g is the gravitation force at angle θ
R dθ is distance traveled by the snowball through angle dθ
sin(θ) represents the downward component of the snowball's trajectory
Plugging in the function m(θ) and rearranging
dW(θ) = g R m0 sin(θ) e^(-α cos(θ)) dθ
W is obtained by integrating the above.
The indefinite integral is
W(θ) = (g R m0 / α) e^(-α cos(θ))
The work performed by the time the snowball reaches the point P is the definite
integral between 0 and Θ, which is
(g R m0 / α) (e^(-α cos(Θ)) - e^(-α cos(0))) = (g R m0 / α) (e^(-α cos(Θ)) - e^-a)