You throw a glob of putty straight up toward the ceiling, which is 3.50 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.10 m/s. 1. What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units.
2. How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.

Explanation:

Given that:

the putty initial speed (u) = 9.10 m/s

distance (s) between hand and the ceiling = 3.50 m

the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.

v² - u² = 2as

Since the putty is moving in a vertical motion(i.e. in an upward direction)

v² - u² = -2gs

v² = u² - 2gs

v = 3.77 m/s

2.

The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:

v = u + at

In an upward direction

v = u - gt

making time t the subject;

v - u = -gt

t = 0.54s

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.