a) How long is the jumper in the air before returning to the Earth?
b) How far does the jumper jump?
The long-jumper is in the air for approximately 0.83 seconds and jumps a distance of approximately 9.14 meters.
Calculation of Time in Air and Jump Distance
Explanation:
To calculate the time in the air and the horizontal distance covered by the long-jumper, we can use the equations of projectile motion.
First, let's find the horizontal and vertical components of the initial velocity:
The horizontal component of the initial velocity (Vx) can be calculated using Vx = V * cos(theta), where V is the initial velocity (12 m/s) and theta is the launch angle (20°).
The vertical component of the initial velocity (Vy) can be calculated using Vy = V * sin(theta), where V is the initial velocity (12 m/s) and theta is the launch angle (20°).
Next, let's calculate the time in the air:
The time in the air (t) can be calculated using the equation t = 2 * Vy / g, where Vy is the vertical component of the initial velocity and g is the acceleration due to gravity (9.8 m/s²).
Now, let's calculate the horizontal distance covered:
The horizontal distance covered (d) can be calculated using the equation d = Vx * t, where Vx is the horizontal component of the initial velocity and t is the time in the air.
Using the given values:
V = 12 m/s
theta = 20°
g = 9.8 m/s²
Let's calculate the time in the air:
Vy = V * sin(theta) = 12 * sin(20°) ≈ 4.08 m/s
t = 2 * Vy / g = 2 * 4.08 / 9.8 ≈ 0.83 seconds
Now, let's calculate the horizontal distance covered:
Vx = V * cos(theta) = 12 * cos(20°) ≈ 11.01 m/s
d = Vx * t = 11.01 * 0.83 ≈ 9.14 meters
This calculation shows that the long-jumper spends approximately 0.83 seconds in the air and jumps a distance of about 9.14 meters. The launch angle and initial velocity play crucial roles in determining these values for the long-jump event.