A baseball is hit a ground level with a velocity of 46m/s at an angle of 30 degrees with the horizontal. The ball lands at ground level on the other side of the fence.
A)how long is the ball in the air?
B)How high does the ball go?
C)How far did the ball land from its starting point?
initial vertical speed is Vi = 46 sin 30 = 23 m/s
v = Vi - 9.81 t
v = 0 at top
so t to top = 23/9.81 = 2.344 s
so total time in air = 4.688 s
H = Vi t - 4.9 t^2
= 23 (2.344) - 4.9(2.344)^2
= 53.92 - 26.69
= 27.2 meters
horizontal speed = 46 cos 30 = 39.8
times time in air 4.69 = 187 meters range
To solve this problem, we can break it down into three parts:
A) Find the time of flight.
B) Find the maximum height.
C) Find the horizontal distance traveled.
A) To find the time of flight, we can use the equation:
time = (2 * initial velocity * sin(angle)) / gravitational acceleration
Substituting the given values:
time = (2 * 46 m/s * sin(30°)) / 9.8 m/s^2
= (92 m/s * 0.5) / 9.8 m/s^2
= 4.7 seconds
Therefore, the ball is in the air for approximately 4.7 seconds.
B) To find the maximum height, we can use the equation for vertical displacement:
maximum height = (initial velocity * sin(angle))^2 / (2 * gravitational acceleration)
Substituting the given values:
maximum height = (46 m/s * sin(30°))^2 / (2 * 9.8 m/s^2)
= (23 m/s * 0.5)^2 / 19.6 m/s^2
= 11.5^2 / 19.6
= 6.782 m
Therefore, the ball reaches a maximum height of approximately 6.782 meters.
C) To find the horizontal distance traveled, we can use the equation:
horizontal distance = initial velocity * cos(angle) * time
Substituting the given values:
horizontal distance = 46 m/s * cos(30°) * 4.7 s
= 46 m/s * 0.866 * 4.7 s
= 181.598 m
Therefore, the ball lands approximately 181.598 meters from its starting point.
To solve this problem, we can use the equations of motion for projectile motion. There are certain key components we need to consider: the initial velocity, the launch angle, time of flight, maximum height, and range.
A) To find the time of flight of the baseball, we can use the equation:
time = (2 * initial velocity * sin(angle)) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values into the equation:
time = (2 * 46 * sin(30)) / 9.8
Simplifying this equation will give us the value of time, giving us the answer to part A.
B) To find the maximum height reached by the baseball, we can use the equation:
maximum height = (initial velocity^2 * (sin(angle))^2) / (2 * g)
Plugging in the known values:
maximum height = (46^2 * (sin(30))^2) / (2 * 9.8)
Simplifying this equation will give us the value of the maximum height, which will answer part B.
C) To find the horizontal range, we can use the equation:
range = (initial velocity^2 * sin(2 * angle)) / g
Substituting the known values:
range = (46^2 * sin(2 * 30)) / 9.8
Simplifying this equation will give us the value of the horizontal range, which will answer part C.
By using these equations and the known values, you can calculate the time of flight, maximum height, and horizontal range of the baseball.