To find the velocity that the second man must have to catch the ball before it touches the ground, we need to break down the initial velocity of the ball into its horizontal and vertical components.
Given: Initial velocity of the ball (v₀) = 12.0 m/s Launch angle (θ) = 40.00° Distance to the second man (d) = 30.0 m
First, we calculate the horizontal and vertical components of the initial velocity using trigonometry:
Horizontal component (v₀x) = v₀ * cos(θ) Vertical component (v₀y) = v₀ * sin(θ)
v₀x = 12.0 m/s * cos(40.00°) ≈ 9.164 m/s (rounded to three decimal places) v₀y = 12.0 m/s * sin(40.00°) ≈ 7.713 m/s (rounded to three decimal places)
Now, we can analyze the horizontal motion of the ball. The horizontal distance traveled by the ball (x) is given as 30.0 m, and the horizontal component of the ball's velocity remains constant throughout its flight. Therefore, we have:
x = v₀x * t
Solving for time (t):
t = x / v₀x
t = 30.0 m / 9.164 m/s ≈ 3.273 s (rounded to three decimal places)
Next, we analyze the vertical motion of the ball. The vertical displacement of the ball is given by the equation:
y = v₀yt - (1/2) * g * t²
Since the ball starts at ground level and lands on the ground, the vertical displacement (y) is zero. We can solve this equation to find the time (t):
0 = v₀y * t - (1/2) * g * t²
Using the acceleration due to gravity (g) as approximately 9.8 m/s²:
(1/2) * g * t² = v₀y * t
(1/2) * 9.8 m/s² * t² = 7.713 m/s * t
(4.9 m/s²) * t² - (7.713 m/s) * t = 0
Solving this quadratic equation, we find two possible solutions for t: t = 0 (which represents the initial time) and t ≈ 1.577 s (rounded to three decimal places).
Since we're interested in the time it takes for the ball to reach the second man, we discard the t = 0 solution.
Now that we have the time it takes for the ball to reach the second man (approximately 1.577 s), we can find the velocity (v₂) the second man must have to catch the ball. We can use the horizontal distance (x) and the time (t) in the horizontal direction:
v₂ = x / t
v₂ = 30.0 m / 1.577 s ≈ 19.022 m/s (rounded to three decimal places)
Therefore, the second man must have a velocity of approximately 19.022 m/s to catch the ball before it touches the ground.