Answer:
1). [tex]R = \frac{(b^2-\frac{2∝}{mv^2_0} )}{sin^2(V_0^2b^2-\frac{2∝}{m})^{1/2}t }[/tex]
where a = ∝
2). [tex]tan( \alpha__l}) = \frac{sin \alpha cm }{cos\alphacm+\frac{V__cm}{V__0} }[/tex]
where ∝ = θ for the above equation
Explanation:
The attached document below clearly depicts the full explanation of the above answers.