Helen Lewis

2022-01-05

A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Robert Pina

Step 1
According to kinematic equation the final velocity is given by
1) ${v}_{y}^{2}={v}_{i}^{2}+2g\mathrm{\Delta }y$
Where it is given that:
(1) The max point so the final velocity ${v}_{y}=0$
(2) The acceleration is $g=-9.8m.{s}^{-2}$
(3) The initial displacement ${y}_{i}=300m$
So substitution in (1) yields
$0={10}^{2}-2×9.8\left(y-300\right)⇒{y}_{m}=305m$
Step 2
The displacement y is given by
2) $y={y}_{i}+{v}_{i}t+\frac{1}{2}{gt}^{2}$
Where it is given that:
(1) The times $t=4s$
(2) The acceleration $g=-9.8m.{s}^{-2}$
(3) The initial velocity ${v}_{i}=10\frac{m}{s}$
(4) The initial displacement ${y}_{i}=300m$
So substitution in (2) yields
$y=300+10\left(4\right)-\frac{1}{2}\left(9.8\right)\left(16\right)⇒y=261.6m$
Which means that the coin is falling down.
Step 3
The time taken before hitting the land is given by equation (2) where it is given that
(1) The final velocity $y=0$
(2) The acceleration $g=-9.8m.{s}^{-2}$
(3) The initial velocity ${v}_{i}=10\frac{m}{s}$
(4) The initial displacement ${y}_{i}=300m$
So substitution in (2) yields
$0=300+10t-\frac{1}{2}×9.8{t}^{2}$
Solving (3) for t and get the positive root
$t=8.9s$

godsrvnt0706

Step 1
General formula for position is:
$x={x}_{0}+{v}_{0}\cdot t+\frac{a\cdot {t}^{2}}{2}$
Where ${x}_{0}$ is starting distance, ${v}_{0}$ is starting velocity and a is acceleration. For this, and $a=-g=-9.8\frac{m}{{s}^{2}}$, because acceleration is directed downwards
Step 2
a) To find maximum height reached, one can derive $x\left(t\right)$ by t and get:
$\frac{d}{dx}x\left(t\right)={v}_{0}+at$
Extremum of function demands first derivative to be equal to zero, thus:
${v}_{0}+at=0;$
$t=-\frac{{v}_{0}}{a}$
$=-\frac{10\frac{m}{s}}{9.8\frac{m}{{s}^{2}}}\approx 1.02s$
Maximum height will be reached at $t=1.02s$ and will be equal to:
$x\left(t=1.02s\right)=300+10\cdot 1.02-\frac{9.8\cdot {1.02}^{2}}{2}=305.102=305.1\left(m\right)$
Step 3
b) Velocity after 4.00 s after release can be calculated as follows:
$v\left(t=4.00s\right)={v}_{0}+at=10-9.8\cdot 4=-29.2\left(\frac{m}{s}\right)$
negative sign means coin is travels downwards. Position after 4.00 s after release:
$x\left(t=4.00s\right)={x}_{0}+{v}_{0}t+\frac{a\cdot {t}^{2}}{2}=300+10\cdot 4.00-\frac{9.8\cdot {4}^{2}}{2}=261.6\left(m\right)$
above the ground.
Step 4
c) Coin hitting the ground means it’s position is 0 m. To get time, one can solve the equation:
$0={x}_{0}+{v}_{0}t+\frac{a\cdot {t}^{2}}{2}$
$D={b}^{2}-4ax={v}_{0}^{2}-4\cdot \frac{a\cdot {x}_{0}}{2}={v}_{0}^{2}-2a{x}_{0}$
$=100-2\cdot \left(-9.8\right)\cdot 300=5980$
$t=\frac{-{v}_{0}±\sqrt{D}}{a}=\frac{-10±77.33}{9.8}$
There are two roots, one of which is negative – but because time can’t be negative, that root is not physical one, so it’s neglected. The one that’s left:
$t=6.87s$

karton

Step 1
Initial position: ${y}_{1}=300m$
Initial velocity: ${v}_{i}=10.0\frac{m}{s}$
Gravity's acceleration: $g=-9.8\frac{m}{{s}^{2}}$
a) From Kinematics: ${v}^{2}={v}_{0}^{2}+2g\mathrm{\Delta }y$
$⇒\mathrm{\Delta }y=\frac{{v}^{2}-{v}_{0}^{2}}{2g}$
When the coin reaches it's maximum height, it's velocity will be $v=0\frac{m}{s}$
$⇒\mathrm{\Delta }y=\frac{{v}_{0}^{2}}{2g}⇒\mathrm{\Delta }y=5.10m$
Maximum height will be $h=300+\mathrm{\Delta }y=305.10m$
Step 2
b) From Kinematics, the movement law will be:
$y\left(t\right)={y}_{0}+{v}_{0}t+\frac{1}{2}g{t}^{2}$
where ${y}_{0}=300m$ is the position where the coin is dropped.
At t=4s
$y\left(4s\right)=261.6m$
c) Again, we have that:
$y\left(t\right)={y}_{0}+{v}_{0}t+\frac{1}{2}g{t}^{2}$
Now, if y final is $y\left({t}_{f}\right)=0$, then
${y}_{0}+{v}_{0}{t}_{f}+\frac{1}{2}g{t}_{f}^{2}=0$
We need to use the resolvent to solve this equation:
$t=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}=\frac{-{v}_{0}±\sqrt{{v}_{0}^{2}}-2g{y}_{0}}{g}$
$⇒t=6.87s$

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