To find the direction and magnitude of the initial velocity of the ball, we can use the kinematic equations of motion and the given information.
Given: Horizontal distance traveled (range) = 60.0 m Time of flight = 5.00 s
Let's consider the horizontal and vertical components of the motion separately.
Horizontal Motion: The horizontal velocity remains constant throughout the motion. The horizontal distance traveled (range) is given as 60.0 m, and the time of flight is 5.00 s. Therefore, we can calculate the horizontal velocity (v₀x) using the formula:
Range = v₀x * time 60.0 m = v₀x * 5.00 s
Solving for v₀x: v₀x = 60.0 m / 5.00 s v₀x = 12.0 m/s
The horizontal velocity (v₀x) is 12.0 m/s.
Vertical Motion: We can analyze the vertical motion using the kinematic equation:
Range = v₀y * time + (1/2) * g * time²
Since the ball lands at the same elevation it was thrown from, the vertical displacement is zero. Hence, the equation becomes:
0 = v₀y * 5.00 s + (1/2) * (-9.8 m/s²) * (5.00 s)²
Simplifying the equation: 0 = v₀y * 5.00 s - 122.5 m
Solving for v₀y: v₀y = 122.5 m / 5.00 s v₀y = 24.5 m/s
The vertical velocity (v₀y) is 24.5 m/s.
Now, we have the horizontal (v₀x = 12.0 m/s) and vertical (v₀y = 24.5 m/s) components of the initial velocity.
To find the magnitude and direction of the initial velocity (v₀), we can use the Pythagorean theorem and trigonometry.
Magnitude of the initial velocity (v₀): v₀ = √(v₀x² + v₀y²) v₀ = √((12.0 m/s)² + (24.5 m/s)²) v₀ ≈ 27.4 m/s
The magnitude of the initial velocity is approximately 27.4 m/s.
Direction of the initial velocity: θ = atan(v₀y / v₀x) θ = atan(24.5 m/s / 12.0 m/s) θ ≈ 63.4°
The direction of the initial velocity is approximately 63.4° (above the horizontal).
Therefore, the direction of the initial velocity is 63.4° above the horizontal, and the magnitude of the initial velocity is approximately 27.4 m/s.