Participants in a road race take water from
a refreshment station and throw their
empty cups away farther down the course.
If a runner has a forward speed of 6.20 m/s,
how far in advance of a garbage pail
should he release his water cup if the
vertical distance between the lid of the
garbage can and the runner’s point of
Answer is 2.0m
please show steps and explain thank you
To calculate the distance the runner should release his water cup in advance of the garbage pail, we need to find the horizontal distance traveled by the runner during the time it takes for the cup to fall to the ground.
Step 1: Calculate the time it takes for the cup to fall to the ground.
The vertical distance between the lid of the garbage can and the runner's point is 2.0 m. We can use the formula for the time of flight of an object in free fall to find the time it takes for the cup to fall.
Using the formula:
h = (1/2)gt^2
where h = vertical distance and g = acceleration due to gravity (9.8 m/s^2), we can rearrange the formula to solve for time (t):
t = sqrt(2h/g)
Substituting the given values:
t = sqrt(2 * 2.0 m / 9.8 m/s^2)
t = sqrt(0.408163265)
t ≈ 0.639 s
Step 2: Calculate the horizontal distance traveled by the runner during this time.
The runner's forward speed is given as 6.20 m/s. We can use the formula for distance traveled to find the horizontal distance covered during the time it took for the cup to fall.
Using the formula:
s = vt
where s = distance, v = velocity, and t = time, we can substitute the values:
s = 6.20 m/s * 0.639 s
s ≈ 3.96 m
Therefore, the runner should release his water cup approximately 3.96 meters in advance of the garbage pail.
To determine the distance in advance a runner should release their water cup, we can use the concept of projectile motion and the kinematic equations of motion.
Assumptions made:
1. There is no air resistance.
2. The water cup is released horizontally.
3. The vertical distance between the lid of the garbage can and the runner's point is 2.0 meters.
First, let's analyze the vertical motion of the water cup. The only force acting on the cup in the vertical direction is gravity. The initial vertical velocity of the cup is zero since it is released horizontally. The vertical displacement of the cup is 2.0 meters, and we need to find the time it takes for the cup to fall that distance.
Using the equation of motion in the vertical direction:
Δy = v₀t + (1/2)gt²
Where:
Δy is the vertical displacement (2.0m),
v₀ is the initial vertical velocity (0 m/s),
g is the acceleration due to gravity (-9.8 m/s²),
t is the time.
Plugging in the values:
2.0 = (0)t + (1/2)(-9.8)t²
Simplifying the equation:
2.0 = (1/2)(-9.8)t²
Rearranging the equation:
4.0 = -4.9t²
Dividing by -4.9:
t² = -0.816
Taking the square root to find t:
t ≈ √(-0.816)
Since time cannot be negative, we can conclude that there is no real solution to this equation. This means it is not physically possible for the cup to fall directly into the garbage can.
Therefore, if the vertical distance remains 2.0 meters, the runner should release their water cup slightly in advance of the garbage can to account for the horizontal distance it will travel during its fall. The exact distance in advance will depend on other factors such as the horizontal speed and the specific trajectory of the cup.
However, without additional information about the horizontal speed, it is not possible to determine the exact distance in advance.
... if the vertical distance between the lid of the garbage can andthe runner’s point of release is 0.50m
h=gt²/2
t=sqrt(2h/g)=
sqrt(2•0.5/9.8) = 0.32 s
s=vt= 6.2•0.32 = 2 m