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?

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°

y-1700=200(x+4)

step-by-step explanation:

200(x+4) is 200x + 800 she is saving 200 per week x is the weeks and the -1700 is. us that’s howmuch she already had

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09