Question

For which angle of projection range of projectile is maximum

(a) 40°
(b) 35°
(e) 43°
(d) 48°
​

Answer 1

the correct answer is e

Step-by-step explanation:

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

R =
`(v_o^2 \ sin 2 \theta )/(g)`

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