1. A construction worker tosses a hammer straight up to a co-worker who is working on a platform above him. The hammer leaves his hand at 7.26 m/sec, and travels 3.33 meters upward, at which point it is caught by the worker on the platform. (It is caught while still moving upward.)
a) What is the acceleration of the hammer after leaving the first worker’s hand and before being caught by the second worker?
b) What is the velocity of the hammer at the instant that it is caught?
c) How long did it take for the hammer to go from the thrower to the catcher?
a. acceleration is due to gravity, g in the downward direction
b. the velocity caught..
vcaught^2=voriginal^2 -2*g*heightabove
solve for vcaught.
c. avg velociy= (vorig+vfinal)/2
time=distance/avgvelocity
To answer these questions, we need to use the equations of motion. Specifically, we can use the kinematic equations:
1) v = u + at,
2) v^2 = u^2 + 2as,
3) s = ut + (1/2)at^2,
where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration,
- s is the displacement, and
- t is the time.
a) To find the acceleration of the hammer, we can use the equation v = u + at. We know the initial velocity (u = 7.26 m/s), the final velocity (v = 0 m/s since the hammer is caught while still moving upward), and we need to find the acceleration (a). Rearranging the equation, we have a = (v - u) / t.
b) To find the velocity of the hammer at the instant it is caught, we can use the equation v = u + at, where the initial velocity (u) and acceleration (a) are the same as in part a). Here, we need to find the final velocity (v).
c) To find the time it took for the hammer to travel from the thrower to the catcher, we can use the equation s = ut + (1/2)at^2. We know the displacement (s = 3.33 m), the initial velocity (u), the acceleration (a), and we need to find the time (t). Rearranging the equation, we have t = (v - u) / a.
Now, let's solve each part step by step:
a) To find the acceleration:
- a = (v - u) / t
- Since the final velocity (v) is 0 m/s (hammer is caught while still moving upward), we have a = (0 - 7.26) / t
- The displacement (s) is not needed for this calculation, so we can skip it.
b) To find the velocity at the instant it is caught:
- v = u + at
- Substituting the given values and acceleration from part a), we have: v = 7.26 + (-a) * t
- Again, displacement (s) is not needed for this calculation, so we can skip it.
c) To find the time taken by the hammer:
- t = (v - u) / a
- Plug in the values of v, u, and a from part a), we have t = (0 - 7.26) / a
- Once again, disregard the displacement (s) since it is not required for this calculation.
If you provide the value of acceleration (a), we can calculate all three values for you.