Question

You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/s. (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling?

Answer 1

Step-by-step explanation:

Given

Height of ceiling is
`h=3.6\ m`

Initial speed of Putty
`u=9.5\ m/s`

Speed of Putty just before it strike the ceiling is given by

`v^2-u^2=2as`

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

`v^2-9.5^2=2* (-9.8)* 3.6`

`v^2=19.69`

`v=4.43\ m/s`

time taken by putty to reach the ceiling

`v=u+at`

`4.43=9.5-9.8* t`

`t=(5.07)/(9.8)`

`t=0.517\ s`