To determine the initial vertical and horizontal velocities of the ball, we can analyze its motion using basic principles of projectile motion.
Given:
- Range (horizontal distance): R = 100 m
- Launch angle (angle with respect to the horizontal): θ = 45 degrees
We need to find the initial vertical velocity (V₀y) and the initial horizontal velocity (V₀x).
The horizontal and vertical components of the initial velocity can be calculated using the following equations:
V₀x = V₀ * cos(θ) V₀y = V₀ * sin(θ)
where V₀ is the initial velocity magnitude, and cos(θ) and sin(θ) are the cosine and sine functions of the angle θ, respectively.
Since the ball reaches the ground, we know that its vertical displacement (Δy) is -100 m (negative because it moves downward). We can use the equation of motion for vertical displacement to find the initial vertical velocity:
Δy = V₀y * t + (1/2) * g * t²
where t is the time of flight and g is the acceleration due to gravity (-9.8 m/s²).
We can rearrange this equation to solve for V₀y:
V₀y = (Δy - (1/2) * g * t²) / t
The time of flight can be determined from the horizontal displacement using the equation:
R = V₀x * t
Rearranging for t:
t = R / V₀x
Substituting this value of t back into the equation for V₀y, we can solve for V₀y:
V₀y = (Δy - (1/2) * g * (R / V₀x)²) / (R / V₀x)
Now, we have an expression for V₀y in terms of V₀x. To find V₀x, we can use the fact that the horizontal and vertical components of the initial velocity are related:
V₀x / V₀y = tan(θ)
Substituting the value of V₀y from the previous equation and rearranging, we can solve for V₀x:
V₀x = (R * g) / [2 * (Δy - R * tan(θ))]
Let's calculate the values.
Given values: R = 100 m θ = 45 degrees g = 9.8 m/s²
Calculations:
V₀x = (R * g) / [2 * (Δy - R * tan(θ))] = (100 * 9.8) / [2 * (-100 - 100 * tan(45))] = 98 / [-2 - 100] = -0.98 m/s
V₀y = (Δy - (1/2) * g * (R / V₀x)²) / (R / V₀x) = (-100 - (1/2) * 9.8 * (100 / -0.98)²) / (100 / -0.98) = (-100 - (1/2) * 9.8 * 102.04) / (100 / -0.98) = (-100 - 501) / (-1) = 601 m/s
Therefore, the initial vertical velocity (V₀y) of the ball is approximately 601 m/s upwards, and the initial horizontal velocity (V₀x) is approximately -0.98 m/s (negative indicates the motion is towards the left).