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>
Answer: just do the same thing, but the problems are different
Explanation: try you best
The train’s average speed is 80km/h
Answer:
The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.
Explanation:
Explanation:
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