A ball is tied to one end of a string. (Projectile Motion)

at given angle suppose it has tangential velocity v

then horizontal component = v*cos(180-θ) = -vcos(θ)

vertical component = vsin(180-θ) = vsin(θ)

Rcos(θ-90) = -vcos(θ)*T --> Rsin(θ)=-vcos(θ)*T

Rsin(θ-90) = vsin(θ)T+.5gT^2 ---> -Rcos(θ)=vsin(θ)T+.5gT^2

put T from the first equation into second

top it has KE = .5mRg
if it is at angle as shown then it is [R-Rsin(θ-90)] below the top point = R(1+sin(θ))

loss in PE =mgR(1+sin(θ))

KE at this point = .5mRg + mgR(1+sin(θ))

we will get v from this equation

equating v from both and solving will give the answer

I hope this may help you some what