A baseball hit just above the ground leaves the bat at 26 m/s at 45∘ above the horizontal.
How far away does the ball strike the ground?
How much time is the ball in the air?
What is its maximum height?
You can find the equations you need in this article. Just scroll down some
https://en.wikipedia.org/wiki/Trajectory
To find the distance the ball strikes the ground, the time the ball is in the air, and its maximum height, we can use a few principles of projectile motion. Let's break down each question step by step:
1. How far away does the ball strike the ground?
To find the horizontal distance, we need to determine the time it takes for the ball to reach the ground. We can use the fact that the vertical component of velocity at the highest point is zero, which means the time taken to reach the highest point is equal to the time taken to fall back to the ground.
Since we know the initial vertical velocity (26 m/s) and the angle above horizontal (45°), we can decompose the vertical velocity using trigonometry:
Vertical Velocity (Vy) = Initial Velocity (V) * sin(Angle)
Vy = 26 m/s * sin(45°) = 18.384 m/s
To calculate the time taken to reach the highest point, we can use the equation:
Time of flight = (2 * Vy) / g
where g is the acceleration due to gravity (9.8 m/s^2).
Time of flight = (2 * 18.384 m/s) / 9.8 m/s^2 ≈ 3.74 s
Now, we can calculate the horizontal distance using the equation:
Horizontal Distance (D) = Vx * Time of flight
where Vx is the horizontal component of velocity.
To find Vx, we can use trigonometry:
Horizontal Velocity (Vx) = Initial Velocity (V) * cos(Angle)
Vx = 26 m/s * cos(45°) ≈ 18.384 m/s
Plugging in the values:
D = 18.384 m/s * 3.74 s ≈ 68.69 m
Therefore, the ball strikes the ground approximately 68.69 meters away.
2. How much time is the ball in the air?
We already calculated the time of flight in the first question, which is approximately 3.74 seconds. Thus, the ball is in the air for about 3.74 seconds.
3. What is its maximum height?
To find the maximum height, we can use the equation for vertical displacement:
Vertical Displacement (d) = (Vy^2) / (2 * g)
Plugging in the values:
d = (18.384 m/s)^2 / (2 * 9.8 m/s^2) ≈ 17.76 m
Therefore, the maximum height of the ball is approximately 17.76 meters.