A baseball batter hits a long fly ball that rises to a height of 49 m. An outfielder on the opposing team can run at 7.7 m/s. What is the farthest the fielder can be from where the ball will land so that it is possible for him to catch the ball?
Well, it seems like the outfielder has some serious competition with gravity here! Let's see if they have what it takes to catch that fly ball.
To figure out the farthest distance the fielder can be from where the ball will land, we need to consider the time it takes for the ball to reach its peak height and then come back down into the outfielder's loving arms.
The time it takes for the ball to reach its peak height can be calculated using the equation:
Time = (Final Velocity - Initial Velocity) / Acceleration
In this case, the initial velocity of the ball is the same as the final velocity when it reaches its peak height since it will be momentarily at rest before starting its descent. Therefore, the formula becomes:
Time = (0 - Initial Velocity) / Acceleration
Now, we can use the equation of motion to find the initial velocity of the ball:
Final Velocity^2 = Initial Velocity^2 + 2 * Acceleration * Distance
Rearranging the formula, we get:
Initial Velocity^2 = Final Velocity^2 - 2 * Acceleration * Distance
Since the final velocity is 0 at the peak height, the equation simplifies to:
Initial Velocity^2 = 2 * Acceleration * Distance
Great! We have everything we need to calculate the initial velocity of the ball.
First, let's determine the acceleration. We'll use the acceleration due to gravity, which is approximately 9.8 m/s^2 (unless gravity has decided to go on vacation, of course!).
Now, let's plug in the numbers and solve for the initial velocity:
Initial Velocity^2 = 2 * 9.8 m/s^2 * 49 m
Initial Velocity^2 = 960.4 m^2/s^2
Initial Velocity ≈ √960.4 m^2/s^2
Initial Velocity ≈ 30.99 m/s (approximately)
So, the initial velocity of the ball is approximately 30.99 m/s.
Now, let's figure out the time it takes for the ball to reach its peak height:
Time = (0 - 30.99 m/s) / (-9.8 m/s^2)
Time ≈ 3.16 s (approximately)
Now that we have the time, we can calculate the distance the fielder can be from where the ball will land:
Distance = Speed * Time
Distance = 7.7 m/s * 3.16 s
Distance ≈ 24.35 m (approximately)
Therefore, the farthest the fielder can be from where the ball will land and still have a chance to catch it is approximately 24.35 meters. That's quite a workout for that outfielder!
To find the farthest distance the fielder can be from where the ball will land, we need to calculate the horizontal distance the ball will travel before reaching the ground. We can use the equations of motion to solve this problem.
First, let's find the time it takes for the ball to reach the ground. We can use the equation for vertical displacement:
y = u * t + (1/2) * a * t^2,
where y is the vertical displacement (49 m), u is the initial vertical velocity (0 m/s), t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).
Plugging in the values, we get:
49 = 0 * t + (1/2) * (-9.8) * t^2.
Simplifying, we have:
4.9t^2 = 49.
Dividing both sides by 4.9, we get:
t^2 = 10.
Taking the square root of both sides, we have:
t ≈ √10.
Therefore, it takes approximately √10 seconds for the ball to reach the ground.
Next, let's calculate the horizontal distance the ball will travel. We can use the equation:
x = v * t,
where x is the horizontal distance, and v is the horizontal velocity of the ball.
Since there are no horizontal forces acting on the ball, the horizontal velocity remains constant throughout the motion. Therefore, we can use the formula:
v = d / t,
where d is the horizontal distance already determined, and t is the time (√10 seconds).
Plugging in the value for d (unknown), and v (unknown), we have:
7.7 = d / √10.
Multiplying both sides by √10, we get:
7.7 * √10 = d.
Evaluating the expression, we have:
d ≈ 24.33.
Therefore, the farthest the fielder can be from where the ball will land is approximately 24.33 meters for it to be possible to catch the ball.
To determine the farthest distance the outfielder can be from where the ball will land, we need to calculate the horizontal distance the ball will travel.
First, let's find the time it takes for the ball to reach its peak height. We can use the following equation:
v = u + at
Where:
v = final velocity (0 m/s at the peak)
u = initial velocity (initial vertical velocity of the ball)
a = acceleration due to gravity (-9.8 m/s²)
t = time
At the peak of the ball's trajectory, the final velocity is 0 m/s, so the equation becomes:
0 = u + (-9.8) × t_peak
Since the ball is launched vertically upward, its initial vertical velocity is:
u = √(2 × g × h)
Where:
g = acceleration due to gravity (9.8 m/s²)
h = maximum height (49 m)
Let's substitute this value into the equation:
0 = √(2 × 9.8 × 49) - 9.8 × t_peak
Squaring both sides of the equation, we get:
0 = 2 × 9.8 × 49 - 9.8² × t_peak²
Simplifying the equation:
t_peak² = 2 × 9.8 × 49 / 9.8²
t_peak² = 98 / 9.8
t_peak² = 10
Therefore, t_peak = √10 ≈ 3.16 seconds.
Now, let's calculate the horizontal distance the ball will travel using the formula:
range = horizontal velocity × time
The horizontal velocity remains constant throughout the ball's flight. To find it, we can divide the horizontal distance by the time of flight:
range = v_horizontal × t_peak
We need to solve for v_horizontal, so let's rearrange the formula:
v_horizontal = range / t_peak
The range is the distance between the batter and the outfielder. We need to find the maximum value for the range so that it is possible for the outfielder to catch the ball.
Therefore, the farthest the fielder can be from where the ball will land is the maximum range of the ball. Since we don't know the horizontal velocity yet, we need additional information to determine the maximum range accurately.