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A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?

Answer :

Answer:

The launch angle should be adjusted to 30.63°

Step-by-step explanation:

The range of a projectile which is the horizontal distance covered by the projectile can be expressed as;

R =(v^2 sin2θ)/g

Where

R = range

v = initial speed

θ = launch angle

g = acceleration due to gravity

For the case above. When the projectile is launched at angle 13° above the horizontal.

θ1 = 13

R1 = (v^2 sin2θ1)/g

R1 = (v^2 sin26°)/g ....1

For the range to double

R2 = (v^2 sin2θ)/g .....2

R2 = 2R1

Substituting R2 and R1

(v^2 sin2θ)/g = 2 × (v^2 sin26°)/g

Divide both sides by v^2/g

sin2θ = 2sin26

2θ = sininverse(2sin26)

θ = sininverse(2sin26)/2

θ = 30.63°

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