An object fired at an angle of 31.9° above the horizontal takes 2.55 s to travel the last 15.5 m of its vertical distance and the last 10.5 m of its horizontal distance. With what speed was the object launched? (Note: This problem does not stipulate that the initial and final elevation of the object has to be the same!)
To find the initial speed at which the object was launched, we can use the kinematic equations of motion. Let's break down the given information:
Angle of launch (θ): 31.9°
Time taken to travel vertical distance (t₁): 2.55 s
Vertical distance traveled (d₁): 15.5 m
Horizontal distance traveled (d₂): 10.5 m
We need to find the initial speed of the object (v₀).
First, let's find the vertical component of the initial velocity (vᵥ) using the equation:
d₁ = vᵥ * t₁ + (1/2) * (-g) * t₁²
where g is the acceleration due to gravity (9.8 m/s²) and -g represents the downward direction.
Plugging in the values:
15.5 = vᵥ * 2.55 + (1/2) * (-9.8) * (2.55)²
Simplifying the equation:
15.5 = vᵥ * 2.55 - 31.4
vᵥ * 2.55 = 15.5 + 31.4
vᵥ * 2.55 = 46.9
vᵥ = 46.9 / 2.55
vᵥ ≈ 18.35 m/s
Now, let's find the horizontal component of the initial velocity (vₕ) using the equation:
d₂ = vₕ * t₁
Plugging in the values:
10.5 = vₕ * 2.55
vₕ = 10.5 / 2.55
vₕ ≈ 4.12 m/s
The initial speed (v₀) can be calculated using the vertical (vᵥ) and horizontal (vₕ) components of velocity:
v₀ = √(vᵥ² + vₕ²)
v₀ = √(18.35² + 4.12²)
v₀ = √(336.4225 + 16.9744)
v₀ = √353.3969
v₀ ≈ 18.80 m/s
Therefore, the object was launched at a speed of approximately 18.80 m/s.