You throw a glob of putty straight up toward the ceiling, which is 3.60 m...

Lucille Davidson

Lucille Davidson

Answered

2022-01-06

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 & Explanation

Melissa Moore

Melissa Moore

Expert

2022-01-07Added 32 answers

Step 1
For freely falling bodies, the equations for the analysis model particle with constant acceleration also hold true.
The acceleration caused by gravity is the steady acceleration of a body falling freely.(ay=g=9.8ms2).
Since, we took the +ve direction to be upward, the acceleration is -ve downward (ay)
a) To calculate vyf, we will use the following equation:
vyf2=vyi2+2ay(yfyi)
yfyi=3.60m
vyf2=9.502+2(9.8)(3.60)
vyf=4.44ms
Step 2
b) We will use the following equation to determine the time:
vyf=vyi+ayt
4.44=9.509.8t
t=0.52s

kalupunangh

kalupunangh

Expert

2022-01-08Added 29 answers

A. V2=Vo2+2gh=9.5219.63.6=19.69.
V=4.44ms.
B. V=Vo+gt=4.44.
9.59.8t=4.44.
9.8t=5.06.
t=0.533s.
karton

karton

Expert

2022-01-11Added 439 answers

Step 1
Given:
Distance, s=3.60 m
NSK
the initial speed of the putty, u = 9.50 m/s
Acceleration,
Step 2
a) We know that Gravitation acceleration acting downwards, so the value of the acceleration is a negative value
that is f=9.80m/s2
Let us take v be the final velocity,
v2=u2+2fs
Substitute the given value:

v=9.50m/s,s=3.60m,f=9.80m/s2v2=(9.50)22×9.80×3.60v2=90.2570.56v=19.69
therefore, v=4.44m/s
Hence the speed of the putty just before it strikes the ceiling is 4.44m/s

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