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Sagot :
Answer:
the correct answer is e
Explanation:
In projectile launching, the range or distance on the x-axis is given by the expression
R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]
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
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