A coin is dropped from a hot-air balloon that is

Step 1
According to kinematic equation the final velocity is given by
1) ${\displaystyle v_y^2=v_i^2+2g\mathrm{\Delta }y}$
Where it is given that:
(1) The max point so the final velocity ${\displaystyle v_y=0}$
(2) The acceleration is ${\displaystyle g=-9.8m.s^{-2}}$
(3) The initial displacement ${\displaystyle y_i=300m}$
So substitution in (1) yields
${\displaystyle 0={10}^2-2\times 9.8(y-300)\Rightarrow y_m=305m}$
Step 2
The displacement y is given by
2) ${\displaystyle y=y_i+v_{i}t+\frac{1}{2}{gt}^2}$
Where it is given that:
(1) The times ${\displaystyle t=4s}$
(2) The acceleration ${\displaystyle g=-9.8m.s^{-2}}$
(3) The initial velocity ${\displaystyle v_i=10\frac{m}{s}}$
(4) The initial displacement ${\displaystyle y_i=300m}$
So substitution in (2) yields
${\displaystyle y=300+10\left(4\right)-\frac{1}{2}\left(9.8\right)\left(16\right)\Rightarrow y=261.6m}$
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 ${\displaystyle y=0}$
(2) The acceleration ${\displaystyle g=-9.8m.s^{-2}}$
(3) The initial velocity ${\displaystyle v_i=10\frac{m}{s}}$
(4) The initial displacement ${\displaystyle y_i=300m}$
So substitution in (2) yields
${\displaystyle 0=300+10t-\frac{1}{2}\times 9.8t^2}$
Solving (3) for t and get the positive root
${\displaystyle t=8.9s}$

Step 1
General formula for position is:
${\displaystyle x=x_0+v_0\cdot t+\frac{a\cdot t^2}{2}}$
Where ${\displaystyle x_0}$ is starting distance, ${\displaystyle v_0}$ is starting velocity and a is acceleration. For this, ${\displaystyle x_0=300m,v_0=10\frac{m}{s}}$ and ${\displaystyle a=-g=-9.8\frac{m}{s^2}}$, because acceleration is directed downwards
Step 2
a) To find maximum height reached, one can derive ${\displaystyle x\left(t\right)}$ by t and get:
$\frac{d}{dx}x(t)=v_0+at$
Extremum of function demands first derivative to be equal to zero, thus:
${\displaystyle v_0+at=0;}$
${\displaystyle t=-\frac{v_0}{a}}$
${\displaystyle =-\frac{10\frac{m}{s}}{9.8\frac{m}{s^2}}\approx 1.02s}$
Maximum height will be reached at ${\displaystyle t=1.02s}$ and will be equal to:
${\displaystyle x(t=1.02s)=300+10\cdot 1.02-\frac{9.8\cdot {1.02}^2}{2}=305.102=305.1\left(m\right)}$
Step 3
b) Velocity after 4.00 s after release can be calculated as follows:
${\displaystyle v(t=4.00s)=v_0+at=10-9.8\cdot 4=-29.2\left(\frac{m}{s}\right)}$
negative sign means coin is travels downwards. Position after 4.00 s after release:
${\displaystyle x(t=4.00s)=x_0+v_{0}t+\frac{a\cdot t^2}{2}=300+10\cdot 4.00-\frac{9.8\cdot 4^2}{2}=261.6\left(m\right)}$
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:
${\displaystyle 0=x_0+v_{0}t+\frac{a\cdot t^2}{2}}$
${\displaystyle D=b^2-4ax=v_0^2-4\cdot \frac{a\cdot x_0}{2}=v_0^2-2a{x}_0}$
${\displaystyle =100-2\cdot (-9.8)\cdot 300=5980}$
${\displaystyle t=\frac{-v_0\pm \sqrt{D}}{a}=\frac{-10\pm 77.33}{9.8}}$
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:
${\displaystyle t=6.87s}$

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: $v^2=v_0^2+2g\mathrm{\Delta }y$
$\Rightarrow \mathrm{\Delta }y=\frac{v^2-v_0^2}{2g}$
When the coin reaches it's maximum height, it's velocity will be $v=0\frac{m}{s}$
$\Rightarrow \mathrm{\Delta }y=\frac{v_0^2}{2g}\Rightarrow \mathrm{\Delta }y=5.10m$
Maximum height will be $h=300+\mathrm{\Delta }y=305.10m$
Step 2
b) From Kinematics, the movement law will be:
$y(t)=y_0+v_{0}t+\frac{1}{2}g{t}^2$
where $y_0=300m$ is the position where the coin is dropped.
At t=4s
$y(4s)=261.6m$
c) Again, we have that:
$y(t)=y_0+v_{0}t+\frac{1}{2}g{t}^2$
Now, if y final is $y(t_f)=0$, 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:
$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-v_0\pm \sqrt{v_0^2}-2g{y}_0}{g}$
$\Rightarrow t=6.87s$