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°
Divide by 6 is the same as times 1/6
-2 and 1/2=-10/2-1/2=-11/2
-11/2 times 1/6=-11/12
Answer:
the slope is 0
Step-by-step explanation:
Answer:
add the y coordinates of the two lines and divide by 2
Step-by-step explanation:
½[0+(-12)]
½[0-12]
½[-12)
y co-ordinate of the midpoint is -6
