How High Does the Flowerpot Rise and How Long Must the Runner Accelerate?

[Matthew]

Hey can someone help me out with one or both of these questions?

1. A dog sees a flowerpot sail up and then back down a window (height of window is 1.77m). If the total time the pot is in sight for 1.0 s, find the height above the top of the window to which the pot rises.


2. A runner hopes to complete a 10-km run in 30.0 min. After exactly 27.0 min there is still 1100 m to go. The runner must then accelerate at 0.20 m/s^2 for how many seconds in order to exactly achieve his desired time?

Thanks in advance

 

[Chi Meson]

Here's some clues for question 1:

This problem is a "two-parter" . The magnitudes of the velocities are the same as the pot passes the top of the window on the way up and one the way down. Ditto for the velocities at the bottom. THis means that you can cut the time given in half and solve for a single "free-fall" problem using displacement, time and acceleration as your known variables, and solve for the velocity at the top of the window. THen do the second part.

 

[M98Ranger]

!

Sure, I'd be happy to help with your questions!

1. To find the height above the top of the window to which the pot rises, we can use the equation of motion for free fall: h = 1/2gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. In this case, we know that the total time the pot is in sight is 1.0 s, so we can plug that in for t. We also know that the height of the window is 1.77 m, so we can plug that in for h. This leaves us with the equation 1.77 = 1/2(9.8)(1.0)^2. Solving for the unknown height, we get h = 1.77 m. Therefore, the height above the top of the window to which the pot rises is 1.77 m.

2. To determine how long the runner needs to accelerate for, we can use the equation d = v0t + 1/2at^2, where d is the distance, v0 is the initial velocity, a is the acceleration, and t is the time. We know that the runner has 1100 m left to go, so we can plug that in for d. We also know that the runner has been running for 27.0 min (1620 s), so we can plug that in for t. The initial velocity is 0 since the runner starts from rest. We can then solve for a using the given acceleration of 0.20 m/s^2, which gives us a = 0.68 m/s^2. Finally, we can plug in all of these values to solve for t: 1100 = (0)(1620) + 1/2(0.68)(t)^2. Solving for t, we get t = 57.7 s. Therefore, the runner needs to accelerate for 57.7 seconds to achieve their desired time of 30.0 min.