A baseball is hit at 29.0 m/s at an angle of 50.0° with the horizontal. Immediately an outfielder runs 4.90 m/s toward the infield and catches the ball at the same height it was hit. What was the original distance between the batter and the outfielder?
To find the original distance between the batter and the outfielder, you can use the kinematic equation for the horizontal motion.
First, let's split the initial velocity of the baseball into its horizontal and vertical components. The horizontal component (Vx) is given by V * cos(θ), where V is the initial velocity of the baseball and θ is the angle it was hit at.
Vx = 29.0 m/s * cos(50.0°)
= 29.0 m/s * 0.6428
= 18.7352 m/s
Next, let's determine the time it takes for the baseball to reach the outfielder. We can use the equation Δx = V * t, where Δx is the displacement, V is the velocity, and t is the time. The displacement is the original distance between the batter and the outfielder, which we are trying to find.
Δx = 18.7352 m/s * t
Since the outfielder is running toward the infield, his speed in the horizontal direction is subtracted from that of the ball. Therefore, the equation becomes:
Δx - 4.90 m/s * t = 0
Now, let's examine the vertical motion of the baseball. The vertical component of the initial velocity (Vy) is given by V * sin(θ), where V is the initial velocity of the baseball and θ is the angle it was hit at.
Vy = 29.0 m/s * sin(50.0°)
= 29.0 m/s * 0.7660
= 22.2940 m/s
Since the outfielder catches the ball at the same height it was hit, the time it takes for the baseball to fall in the vertical direction is the same as the time it takes for the outfielder to reach the ball. Therefore, we can use the time t from the equation above to calculate the vertical distance traveled by the baseball:
Δy = Vy * t
= 22.2940 m/s * t
Since the initial vertical velocity of the ball is positive and the ball is caught at the same height it was hit, the vertical displacement, Δy, is zero. This allows us to write the equation:
22.2940 m/s * t - (1/2) * 9.8 m/s^2 * t^2 = 0
Now, let's solve this quadratic equation for t.
(1/2) * 9.8 m/s^2 * t^2 - 22.2940 m/s * t = 0
Multiplying through by 2 to get rid of the fraction:
9.8 m/s^2 * t^2 - 44.5880 m/s * t = 0
This equation can be factored as:
t(9.8 m/s * t - 44.5880 m/s) = 0
This gives us two possible solutions: t = 0 (which we discard as it represents initial conditions) or:
9.8 m/s * t - 44.5880 m/s = 0
9.8 m/s * t = 44.5880 m/s
t = 44.5880 m/s / 9.8 m/s
t ≈ 4.55 s
Now, let's use this value of t in the equation for horizontal motion to find the original distance between the batter and the outfielder:
Δx = 18.7352 m/s * t
= 18.7352 m/s * 4.55 s
≈ 85.115 m
Therefore, the original distance between the batter and the outfielder is approximately 85.115 meters.
To find the original distance between the batter and the outfielder, we need to break down the motion of the ball into its horizontal and vertical components.
First, let's find the time it takes for the outfielder to catch the ball. Since the vertical displacement is zero (the ball was caught at the same height it was hit), we can use the equations of motion to determine the time. The equation we'll use is:
Δy = V₀y * t + (1/2) * g * t²
Since the initial vertical velocity is given by V₀y = V₀ * sin(θ), where V₀ is the initial speed of the ball and θ is the launch angle, and the acceleration due to gravity is g = -9.8 m/s², we can rewrite the equation as:
0 = (V₀ * sin(θ)) * t + (1/2) * (-9.8) * t²
Simplifying the equation, we get:
4.9 * t² = (V₀ * sin(θ)) * t
Since we want to find the time t, we divide both sides of the equation by (V₀ * sin(θ)):
4.9 * t = V₀ * sin(θ)
Now, we can solve for t:
t = (V₀ * sin(θ)) / 4.9
Next, we can find the horizontal distance traveled by the ball using the equation:
Δx = V₀x * t
where V₀x is the initial horizontal velocity, given by V₀x = V₀ * cos(θ). By substituting the known values, we get:
Δx = (V₀ * cos(θ)) * ((V₀ * sin(θ)) / 4.9)
Finally, we can calculate the original distance between the batter and the outfielder by considering the outfielder's velocity. The distance traveled by the outfielder in the same time interval is given by:
distance = velocity * time
where the velocity is the speed of the outfielder, given as 4.90 m/s, and the time is given by t.
Therefore, the original distance between the batter and the outfielder is:
distance = 4.90 * ((V₀ * sin(θ)) / 4.9) = V₀ * sin(θ)
Substituting the given values, we have:
distance = 29.0 * sin(50.0°)
Calculating this expression, we find:
distance ≈ 22.1 meters
So, the original distance between the batter and the outfielder was approximately 22.1 meters.