A pitching machine delivers a ball. if the ball is launched with an initial velocity of 22 m/s^2 [30d above horizontal] and the player hits it at the same height from which it was launched.
a) how long is the baseball in the air on its way to the batter?
b) how far away (range) is the pitching machine from the batter?
u = horizontal velocity = 22 cos 30 = 19.05 m/s
d = 19.05 * T
but vertical problem same T
Vi= 22 sin 30 = 11 m/s
h= Hi + Vi T - 4.9 T^2
but h = Hi
4.9 T^2 = 11T
T = 11/4.9 = 2.24 s in air
d = 19.05 * 2.24 = 42.8 meters
To solve this problem, we can break it down into two parts:
a) Finding the time the ball is in the air:
Since the ball is launched at an angle above the horizontal, we need to split the initial velocity into horizontal and vertical components.
The horizontal component is given by: Vx = V * cos(θ), where V is the initial velocity (22 m/s) and θ is the angle (30 degrees).
Vx = 22 m/s * cos(30 degrees) = 22 m/s * √3/2 ≈ 19.04 m/s
The vertical component is given by: Vy = V * sin(θ).
Vy = 22 m/s * sin(30 degrees) = 22 m/s * 1/2 = 11 m/s
Now, the time of flight (t) can be found using the vertical component of motion. We can use the equation: y = Vy * t - (1/2) * g * t^2, where y is the initial height (same as the height at which it was launched), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
We can rewrite the equation as: 0 = Vy * t - (1/2) * g * t^2
Let's solve for t:
(1/2) * g * t^2 = Vy * t
(1/2) * 9.8 * t^2 = 11 * t
4.9 * t^2 - 11 * t = 0
t(4.9 * t - 11) = 0
We have two possibilities:
1) t = 0 (which is not possible because the ball is in the air)
2) 4.9 * t - 11 = 0
4.9 * t = 11
t ≈ 2.24 seconds
Therefore, the ball is in the air for approximately 2.24 seconds.
b) Finding the range:
The range (horizontal distance) can be calculated using the equation: R = Vx * t, where R is the range, Vx is the horizontal component of the initial velocity, and t is the time of flight.
R = 19.04 m/s * 2.24 s ≈ 42.63 meters
Therefore, the pitching machine is approximately 42.63 meters away from the batter.
To answer these questions, we can use the principles of projectile motion. Let's break down each part of the question.
a) How long is the baseball in the air on its way to the batter?
To find the time the baseball is in the air, we need to determine the vertical motion separately from the horizontal motion.
The initial velocity of the ball is provided as 22 m/s at an angle of 30 degrees above the horizontal. We need to find the vertical component of this velocity.
Vertical component of velocity (Vy) = Velocity x sin(angle)
Vy = 22 m/s x sin(30)
Vy = 11 m/s
Now we can use the equation of motion for vertical displacement to find the time of flight.
Vertical displacement (sy) = Vy x time + 0.5 x acceleration x time^2
Since the ball is launched and caught at the same height, the vertical displacement is zero.
0 = 11 m/s x time + 0.5 x (-9.8 m/s^2) x time^2
Now we can solve this quadratic equation to find the time.
0 = 11t - 4.9t^2
We can rearrange the equation to get:
4.9t^2 - 11t = 0
Using factoring, we find:
t(4.9t - 11) = 0
So either t = 0 (which is not the answer) or t = 11/4.9
Thus, the time that the baseball is in the air on its way to the batter is approximately 2.24 seconds.
b) How far away (range) is the pitching machine from the batter?
To find the range, we can use the horizontal component of the initial velocity.
Horizontal component of velocity (Vx) = Velocity x cos(angle)
Vx = 22 m/s x cos(30)
Vx = 19.05 m/s
To find the range, we need to multiply the horizontal component of velocity by the time of flight.
Range (R) = Vx x time
R = 19.05 m/s x 2.24 s
Thus, the pitching machine is approximately 42.72 meters away from the batter.