 Lucille Davidson

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? Melissa Moore

Expert

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.$\left({a}_{y}=g=9.8\frac{m}{{s}^{2}}\right)$.
Since, we took the +ve direction to be upward, the acceleration is -ve downward $\left(-{a}_{y}\right)$
a) To calculate ${v}_{yf}$, we will use the following equation:
${v}_{yf}^{2}={v}_{yi}^{2}+2{a}_{y}\left({y}_{f}-{y}_{i}\right)$
${y}_{f}-{y}_{i}=3.60m$
${v}_{yf}^{2}={9.50}^{2}+2\left(-9.8\right)\left(3.60\right)$
${v}_{yf}=4.44\frac{m}{s}$
Step 2
b) We will use the following equation to determine the time:
${v}_{yf}={v}_{yi}+{a}_{y}t$
$4.44=9.50-9.8t$
$t=0.52s$ kalupunangh

Expert

A. ${V}^{2}=V{o}^{2}+2g\cdot h={9.5}^{2}-19.6\cdot 3.6=19.69$.
$V=4.44\frac{m}{s}$.
B. $V=Vo+g\cdot t=4.44$.
$9.5-9.8t=4.44$.
$-9.8t=-5.06$.
$t=0.533s.$ karton

Expert

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 $f=-9.80m/{s}^{2}$
Let us take v be the final velocity,
${v}^{2}={u}^{2}+2fs$
Substitute the given value:

$\begin{array}{}v=9.50m/s,s=3.60m,f=-9.80m/{s}^{2}\\ {v}^{2}=\left(9.50{\right)}^{2}-2×9.80×3.60\\ {v}^{2}=90.25-70.56\\ v=\sqrt{19.69}\end{array}$
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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