A ball is to be shot from level ground toward a wall at distance x (Figure a). (Figure b) shows the y component vy of the ball’s velocity just as it would reach the wall, as a function of that distance x. The scaling is set by vys= 5.0 m/s and xs = 20 m.What is the launch angle?
The Correct Answer and Explanation is :
To determine the launch angle (θ) of a projectile based on the given graph of the vertical velocity component (vy) as a function of horizontal distance (x), we can analyze the relationship between these variables in projectile motion.
Understanding Projectile Motion:
In projectile motion, the horizontal (vx) and vertical (vy) components of velocity are influenced differently:
- Horizontal Component (vx): Remains constant throughout the flight, as there is no acceleration in the horizontal direction (assuming air resistance is negligible).
- Vertical Component (vy): Changes linearly over time due to the constant acceleration caused by gravity (g ≈ 9.8 m/s² downward).
The vertical velocity at any point can be described by:
vy = v₀y – g * t
where v₀y is the initial vertical velocity component, and t is the time elapsed.
Analyzing the Graph:
The provided graph shows vy as a function of x, with scaling factors vys = 5.0 m/s and xs = 20 m. This indicates that at x = 20 m, vy = 5.0 m/s.
Relating Horizontal Distance and Time:
The horizontal distance traveled is given by:
x = vx * t
Solving for time (t):
t = x / vx
Expressing vy in Terms of x:
Substitute t into the equation for vy:
vy = v₀y – g * (x / vx)
Since vx = v₀ * cos(θ) and v₀y = v₀ * sin(θ), we have:
vy = v₀ * sin(θ) – g * (x / (v₀ * cos(θ)))
Simplifying:
vy = v₀ * sin(θ) – (g * x) / (v₀ * cos(θ))
Using the Given Data:
At x = 20 m, vy = 5.0 m/s. Plugging these values into the equation:
5.0 m/s = v₀ * sin(θ) – (9.8 m/s² * 20 m) / (v₀ * cos(θ))
Rearranging terms:
5.0 m/s = v₀ * sin(θ) – (196 m²/s²) / (v₀ * cos(θ))
Let k = v₀ * cos(θ). Then, v₀ * sin(θ) = k * tan(θ), and the equation becomes:
5.0 m/s = k * tan(θ) – 196 m²/s² / k
Multiplying through by k to clear the denominator:
5.0 m/s * k = k² * tan(θ) – 196 m²/s²
Rearranging:
k² * tan(θ) – 5.0 m/s * k – 196 m²/s² = 0
This is a quadratic equation in terms of k. Solving for k using the quadratic formula:
k = [5.0 m/s ± sqrt((5.0 m/s)² + 4 * tan(θ) * 196 m²/s²)] / (2 * tan(θ))
Since k = v₀ * cos(θ), we can express v₀ as:
v₀ = k / cos(θ)
Determining the Launch Angle (θ):
To find θ, we need to know v₀ or have additional information. However, if we assume that the projectile reaches its maximum height at the wall (x = 20 m), then vy = 0 m/s at that point. This simplifies our earlier equation:
0 = v₀ * sin(θ) – (9.8 m/s² * 20 m) / (v₀ * cos(θ))
Solving for θ:
v₀ * sin(θ) = (9.8 m/s² * 20 m) / (v₀ * cos(θ))
tan(θ) = (9.8 m/s² * 20 m) / v₀²
θ = arctan((9.8 m/s² * 20 m) / v₀²)
Without the exact value of v₀, we cannot determine a numerical value for θ. Additional information, such as the initial speed or the time of flight, would be necessary to calculate the precise launch angle.
In summary, the launch angle θ can be determined from the relationship between vy and x, but specific numerical values require knowledge of the initial velocity or other parameters of the projectile’s motion.