To solve this problem, we can use the equations of motion for projectile motion. Let's consider the second ball thrown at an angle of 30.0° above the horizontal.
We'll assume that there is no air resistance and that the acceleration due to gravity is approximately 9.8 m/s² downward.
The given information is as follows:
- Time of flight (total time taken for the ball to return to its original level): t = 2.75 s
- Angle of projection: θ = 30.0°
To find the initial speed (magnitude) of the second ball, we need to break down its initial velocity into its horizontal and vertical components. We can use the equations:
Horizontal component: Vx = V₀ * cos(θ) Vertical component: Vy = V₀ * sin(θ)
Where V₀ is the initial speed.
Since the ball returns to its original level, we know that the vertical displacement is zero. The formula for the vertical displacement is given by:
Δy = Vy * t + (1/2) * g * t²
Since Δy is zero, we can simplify the equation to:
0 = Vy * t - (1/2) * g * t²
Now, substituting the expressions for Vy and Vx into the equation, we get:
0 = (V₀ * sin(θ)) * t - (1/2) * g * t²
Simplifying further:
0 = (V₀ * sin(θ)) * t - (4.9) * t²
Now, we can solve this quadratic equation for V₀. Rearranging and factoring out t:
0 = t * [(V₀ * sin(θ)) - (4.9) * t]
Since we know t ≠ 0, we can solve the expression in the brackets:
(V₀ * sin(θ)) - (4.9) * t = 0
V₀ * sin(θ) = (4.9) * t
Finally, solving for V₀:
V₀ = (4.9 * t) / sin(θ)
Plugging in the given values:
V₀ = (4.9 * 2.75) / sin(30.0°)
V₀ ≈ 26.95 m/s
Therefore, the initial speed of the second ball is approximately 26.95 m/s.