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? i got a=0

b) What is the velocity of the hammer at the instant that it is caught? i got v= 3.26

c) How long did it take for the hammer to go from the thrower to the catcher? i got t=5

please help.

the acceleration is as usual: -9.8 m/s^2 due to gravity

Considering h=0 when it was thrown, we have

h(t) = 7.26t - 4.9t^2 = 3.3

I don't get your answers. In fact, I suspect a typo. What did you do?

Let's break down the problem step by step to find the answers.

a) To find the acceleration of the hammer, we can use the equation of motion:
v^2 = u^2 + 2as

Here, v is the final velocity (which is the velocity at the instant of catching, and in this case, it's 0), u is the initial velocity (7.26 m/sec), a is the acceleration, and s is the displacement (3.33 meters).

Plugging in the values:
0 = (7.26)^2 + 2a(3.33)

Rearranging the equation, we get:
a = -[(7.26)^2 / (2 * 3.33)]

Simplifying the calculation:
a ≈ -6.855 m/s^2

The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means the hammer is decelerating or slowing down.

b) To find the velocity of the hammer at the instant it is caught, we can use the equation:
v = u + at

Here, v is the final velocity (which we have to find), u is the initial velocity (7.26 m/sec), a is the acceleration (-6.855 m/s^2), and t is the time it takes for the hammer to travel.

Since the hammer is caught while still moving upward, the final velocity is 0. So, the equation becomes:
0 = 7.26 + (-6.855)t

Rearranging the equation to solve for t:
t = 7.26 / 6.855

Calculating the value of t:
t ≈ 1.059 seconds

c) The time it took for the hammer to go from the thrower to the catcher is approximately 1.059 seconds.

Therefore:
a) The acceleration of the hammer is approximately -6.855 m/s^2.
b) The velocity of the hammer at the instant of catching is approximately 0 m/s.
c) The hammer took approximately 1.059 seconds to go from the thrower to the catcher.