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 6.20m/s, how far in advance of a garbage pail should he realease his water cup if the vertical distance between the lid of the garbage can and the runner's point of release is 0.50m?
pls help. When i try to solve this problem i first find the t which is time and i got 0.32seconds and then i used d=vt formula to find the d and I got 1.984m. i do not know if it is right. pls help me.
To solve this problem, you need to consider the horizontal and vertical components of the runner's motion separately.
First, let's calculate the time it takes for the runner to reach the garbage pail. We can use the formula for time, t = d/v, where d is the horizontal distance and v is the runner's forward speed.
Given that the forward speed is 6.20 m/s, and we want to find the time it takes for the runner to reach the garbage pail, we substitute these values into the formula:
t = d/v
t = 0.50 m / 6.20 m/s
t ≈ 0.081 seconds
Now, we need to find the horizontal distance the runner covers in this time. We can use the formula for distance, d = vt, where v is the runner's forward speed and t is the time.
Substituting the values we have, we get:
d = 6.20 m/s × 0.081 s
d ≈ 0.5022 m
Therefore, the runner should release their water cup approximately 0.5022 meters or about 50.22 cm in advance of the garbage pail.