You throw a glob of putty straight up toward the ceiling, which is 3.00m above the point where the putty leaves your hand. the initial speed of the putty as it leaves your hand is 8.40m/s . what is the speed of the putty just before it strikes the ceiling? how much time from when it leaves your hand does it take the putty to reach the ceiling?

It takes the putty 0.507 s to reach the ceiling.

The velocity of the putty just before it strikes the ceiling is 3.43 m/s.

Explanation:

Hi there!

The height of the glob of putty after a time "t" and its velocity are given by the following equations:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height at a time "t".

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at a time "t".

First, let´s find the time it takes the putty to reach the ceiling. We have to find at which time the putty is at a height of 3.00 m from your hand. Let´s place the origin of the frame of reference at your hand so that h0 = 0. The initial velocity, v0, is 8.40 m/s. Then:

h = h0 + v0 · t + 1/2 · g · t²

3.00 m = 0 m + 8.40 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -3.00 m + 8.40 m/s · t  - 4.9 m/s² · t²

Solving the quadratic equation using the quadratic formula with these coefficients:

a = -4.9

b = 8.4

c = -3

t = 0.507 s and t = 1.21 s

We use the value that happens before, that is, t = 0.507 s.

It takes the putty 0.507 s to reach the ceiling.

Now, let´s calculate the velocity of the putty at t = 0.507 s:

v = v0 + g · t

v = 8.40 m/s - 9.8 m/s² · 0.507 s

v = 3.43 m/s

The velocity of the putty just before it strikes the ceiling is 3.43 m/s.

moving ball hits still ball. mover stops, hit ball moves. momentum transfer.