Answer: E

Solution:

Take the first jump as $(2,0)$ and the second as $(2\cos\theta,2\sin\theta)$ with $\theta$ uniform on $[0,2\pi).$ The landing distance satisfies $R^2=(2+2\cos\theta)^2$ $+(2\sin\theta)^2$ $=8+8\cos\theta.$ We need $R\lt 1,$ i.e. $\cos\theta\lt-\tfrac78.$ The measure of such angles is $2\arccos\tfrac78,$ so the probability is $\dfrac{2\arccos\tfrac78}{2\pi}= \dfrac{\arccos\tfrac78}{\pi}.$ Using $\arccos(1-2x^2)=2\arcsin x$ with $x=\tfrac14$ gives $\arccos\tfrac78=2\arcsin\tfrac14,$ so the probability is $\dfrac{2\arcsin\tfrac14}{\pi}.$

Thus, the correct answer is E.