Vo = 26m/s[45o].
Xo = 26*Cos45 = 18.38 m/s. = Hor. comp.
Yo = 26*sin45 = 18.38 m/s. = Ver. comp.
a. Y^2 = Yo^2 + 2g*h.
0 = 18.38^2 + (-19.6)*h, h = ?.
Range = Vo^2*sin(2A)/g.
Range = 26^2*sin(90)/9.8,
b. Range = Xo*T. T = ?.
Xo = 26*Cos45 = 18.38 m/s. = Hor. comp.
Yo = 26*sin45 = 18.38 m/s. = Ver. comp.
a. Y^2 = Yo^2 + 2g*h.
0 = 18.38^2 + (-19.6)*h, h = ?.
Range = Vo^2*sin(2A)/g.
Range = 26^2*sin(90)/9.8,
b. Range = Xo*T. T = ?.
1. Determine the initial vertical and horizontal velocities:
The initial velocity of 26 m/s is at an angle of 45 degrees. To find the vertical and horizontal components, we can use trigonometry:
Vertical velocity (Vy) = initial velocity * sin(angle)
Horizontal velocity (Vx) = initial velocity * cos(angle)
Vy = 26 m/s * sin(45°) = 26 m/s * 0.707 ≈ 18.382 m/s
Vx = 26 m/s * cos(45°) = 26 m/s * 0.707 ≈ 18.382 m/s
2. Find the time of flight:
The time of flight (T) is the total time the discus stays in the air. Considering that the discus reaches its maximum height and lands at the same height, the time of flight is twice the time it takes to reach maximum height.
We can use the equation:
Vy = initial velocity (Vy) - (acceleration * time)
Since the discus reaches its maximum height when Vy becomes zero, we can find the time it takes to reach maximum height by setting Vy to zero and solving for time.
0 = Vy - (acceleration * time)
0 = 18.382 m/s - (9.8 m/s² * time)
Solving for time:
18.382 m/s = 9.8 m/s² * time
time = 18.382 m/s / 9.8 m/s² ≈ 1.875 s
The time of flight (T) is twice this time: T = 2 * 1.875 s = 3.75 s
3. Calculate the maximum height:
To find the maximum height (H), we can use the equation:
H = (Vy^2) / (2 * acceleration)
H = (18.382 m/s)^2 / (2 * 9.8 m/s²)
H ≈ 19 m
4. Determine the range:
The range (R) is the horizontal distance traveled by the discus. We can use the formula:
R = Vx * T
R = 18.382 m/s * 3.75 s
R ≈ 68.922 m
So, the champion discus thrower attains a maximum height of approximately 19 meters, a range of about 68.922 meters, and a time of flight of 3.75 seconds.