Answer:
Step-by-step explanation:
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°
Property of inverse functions:
Therefore the answer to both is True.
20+3=23 divided by 2 ==11.5
Answer:
volume of xy-plane outside the cone = 16π/3
Step-by-step explanation:
using cylindrical coordinates
z² = x² +y² =====>z²=r²=====>z=r
x² + y² =4 ====>r = 2
So, the volume ∫∫∫dV equal
∫(θ = 0 to 2π) ∫(r = 0 to 2) ∫z=0 to r) 1 x (r dz dr dθ) via cylindrical coordinates
= ∫(θ = 0 to 2π) ∫(r = 0 to 2) r² dr dθ
= ∫(θ = 0 to 2π) (1/3)r³ {for r = 0 to 2} dθ
= 2π x 8/3
= 16π/3
