Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
<u>θ₀ = 5.22°</u>
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
<u>θ₀ = 84.78°</u>
The new gravitational attraction will be 1/4 as much
Explanation:
The magnitude of the gravitational force between two objects is given by
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, the original force between the two objects is F, when they are separated by a distance r.
Later, the distance between the two objects is doubled, so the new distance is

Therefore, the new force will be

Therefore, the new force will be one-fourth as much.
Learn more about gravitational force:
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