For which angle of projection range of projectile is maximum (a) 40°
(b) 35°
(e) 43°
(d) 48°
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the correct answer is e

Explanation:

In projectile launching, the range or distance on the x-axis is given by the expression

         R =

where v₀ is the initial velocity θ the launch angle.

When analyzing this expression is maximum when the sine function is maximum

           sin 2 θ = 1

           2θ = sin⁻¹  1

          2 θ = 90

           θ = 90/2

           θ = 45º

reviewing the answers there are no correct ones since the maximum angle occurs for 45º

if we analyze with these angles we have

θ = 40

            sin 2θ = sin 2 40 = 0.98

θ = 35

             sin 2 35 = 0.939

θ = 43

             sin 2 43 = 0.998

θ = 48

             sin 2 48 = 0.994

if we take the written angles the range is maximum for 43º

therefore the correct answer is e

found in the nucleus

has mass of one amu

the answer is 3.1 his answer is wrong mines is right