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:
Part a)

Part b)

Part c)

Explanation:
As we know that acceleration is rate of change in velocity of the object
So here we know that


Part a)
differentiate x and y two times with respect to time to find the acceleration






Now the acceleration of the object is given as

at t= 1.1 s we have

now the net force of the object is given as



now magnitude of the force will be

Part b)
Direction of the force is given as



Part c)
For velocity of the particle we have




now at t = 1.1 s

now the direction of the velocity is given as



Answer:
3.5 seconds of flight time; 13.9 m from the base of the cliff
Explanation:
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