An arrow has an initial launch speed of 38 m/s. If it must strike a target 37 m away at the same elevation, what should be the projection angle?(Hint: sin2θ = 2sinθcosθ)

θ = 14.54°/2 = 7.27°

Therefore, the projection angle θ = 7.27°

Step-by-step explanation:

The range of a projectile can be written as;

R = (v^2 . Sin2θ)/g 1

Where;

R = range = 37m

v = initial speed = 38m/s

θ = projection angle

g = acceleration due to gravity = 9.8m/s^2

Making sin2θ the subject of formula in equation 1

Sin2θ = Rg/v^2

Sin2θ = (37×9.8)/38^2

Sin2θ = 0.2511

2θ = sininverse(0.2511)

2θ = 14.54°

θ = 14.54°/2 = 7.27°

Therefore, the projection angle θ = 7.27°

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