To determine the initial vertical and horizontal velocities of the ball, we can analyze the motion and use the equations of motion in projectile motion.
Given:
- Range (horizontal distance traveled): R = 100 m
- Launch angle: θ = 45 degrees
We can break down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to the effect of gravity.
The equations for projectile motion are as follows:
Horizontal motion:
- Vx = initial horizontal velocity
- Dx = horizontal distance traveled (Range)
- Time of flight (T) = Dx / Vx
Vertical motion:
- Vy = initial vertical velocity
- Dy = vertical distance traveled
- Dy = Vy * t - (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.
To find Vx and Vy, we need to solve for these variables using the given information.
First, let's find the time of flight (T) using the horizontal distance (Range): Dx = Vx * T 100 = Vx * T
Next, let's find the vertical distance traveled (Dy) using the time of flight: Dy = Vy * T - (1/2) * g * T^2 Since the ball reaches the ground, Dy = 0: 0 = Vy * T - (1/2) * g * T^2
Now, we can solve the two equations to find Vx and Vy. Rearranging the first equation for T, we get: T = 100 / Vx
Substituting T in the second equation: 0 = Vy * (100 / Vx) - (1/2) * g * (100 / Vx)^2
Simplifying the equation: 0 = 100Vy / Vx - (g * 100^2) / Vx^2
Rearranging and multiplying through by Vx^2: 0 = 100Vy * Vx - g * 100^2
Finally, we can solve for Vx and Vy: Vx = sqrt(g * 100^2 / (100Vy)) Vy = g * 100^2 / (100Vy)
Using the approximate value of g as 9.8 m/s^2, we can calculate the velocities.
Vx = sqrt((9.8 * 100^2) / (100 * tan(45))) = sqrt(9800) ≈ 99 m/s
Vy = 9.8 * 100^2 / (100 * tan(45)) = 9800 / 1 = 9800 m/s
Therefore, the initial vertical velocity is approximately 9800 m/s, and the initial horizontal velocity is approximately 99 m/s.