$v^{2} = 0 + 2ad$

= $sqrt{2ad}$

= $sqrt{2×9.8×3.7}$

= 8.28m/s

x=3.7m

a=9.8m/s$^2$

=============

V=?

t : Time

g : acceleration

$v_{f}$ : final velocity

d: displacement

– Window height: $h=3.5mathrm{~m}$

We need to find:

– Final velocity $v_f$

Since we are given the distance that the ball travels $h$, and we know that gravity accelerates the ball downwards with acceleration of $gapprox9.8mathrm{~dfrac{m}{s^2}}$, we need to express how velocity changes, depending only on the distance traveled and the acceleration.

$$begin{align}
v=v_0+acdot t
end{align}$$

Where $v_0$ is the initial velocity, meaning $-$ velocity at the $t=0$.

Now we don’t want our equation for velocity to depend on time, since we are not given the time, but rather the distance traveled. Let’s figure out how to do that.

$$begin{align}
S=v_0cdot t+dfrac{acdot t^2}{2}
end{align}$$

Now let’s se how can we merge equations $(2)$ and $(1)$ in order to get the velocity $v$ depending only on acceleration $a$ and distance traveled $S$.

$$begin{align}
v&=v_0+acdot t ~~ /^2 nonumber
\v^2&=v_0^2+2cdot acdot tcdot v_0+(acdot t)^2
end{align}$$

Now we can factor out the term $2cdot a$:

$$begin{align}
v^2&=v_0^2+2cdot acdot(v_0cdot t+dfrac{acdot t^2}{2})
end{align}$$

$$begin{align}
v^2&=v_0^2+2cdot acdot(v_0cdot t+dfrac{acdot t^2}{2}) nonumber
\&=v_0^2+2cdot acdot S
end{align}$$

Now, using $(5)$ we can finally solve our problem.

$$begin{aligned}
v^2&=v_0^2+2cdot acdot S nonumber
\v_f^2&=0+2cdot gcdot S
\&downarrow
\v_f&=sqrt{2cdot gcdot S}
\&=sqrt{2cdot 9.81cdot 3.5}
\&=boxed{8.29mathrm{~dfrac{m}{s^2}}}
end{aligned}$$

– Intitial speed: $v_0=22.5mathrm{~dfrac{m}{s}}$

We need to find:

$mathrm{bold a)}$ Height $h$

$mathrm{bold b)}$ Time of travel $t$

Since we are given initial velocity $v_0$, and are asked to find the height $h$ (which is half of the whole distance traveled) and time of travel $t$ (where half of the time ball spends going up and the other half going down), we need to relate these 3 quantities.

To do this, we can manipulate kinematic equations.

$$begin{align}
v=v_0+acdot t
end{align}$$

Where $v_0$ is the initial velocity, meaning $-$ velocity at the $t=0$.

We can also relate the distance traveled $S$ to the time of travel $t$:

$$begin{align}
S=v_0cdot t+dfrac{acdot t^2}{2}
end{align}$$

We can square the equation $(1)$ and get:

$$begin{align}
v&=v_0+acdot t ~~ /^2 nonumber
\v^2&=v_0^2+2cdot acdot tcdot v_0+(acdot t)^2
end{align}$$

Now we can factor out the term $2cdot a$:

$$begin{align}
v^2&=v_0^2+2cdot acdot(v_0cdot t+dfrac{acdot t^2}{2})
end{align}$$

We can now compare the equations $(3)$ and $(4)$, and see that the second term on the right side of the equation $(4)$ contains $S$ in it self:

$$begin{align}
v^2&=v_0^2+2cdot acdot(v_0cdot t+dfrac{acdot t^2}{2}) nonumber
\&=v_0^2+2cdot acdot S
end{align}$$

Now, by noticing that at the top of the path (at $S=h$), velocity of the ball drops to zero ($v=0$), we can get the height from $(5)$.

$$begin{aligned}
v^2&=v_0^2+2cdot acdot S
\0&=v_0^2+2cdot gcdot h
end{aligned}$$
Thus,
$$begin{aligned}
h=-dfrac{v_0^2}{2cdot g}=boxed{25.8mathrm{~m}}
end{aligned}$$

Where we ignored the minus sign, since the acceleration $g$ is also in opposite direction for ball going up.

We can find the time it takes for ball to go up $t_{up}$ by using equation $(1)$, which is the same time it takes for ball to drop down $t_{down}$. Thus, the time of whole path $t$ is:

$$begin{align}
t=t_{up}+t_{down}=t_{up}+t_{up}
=2cdot t_{up}end{align}$$

Now, let’s find the time $t_{up}$.

$$begin{aligned}
v&=v_0+acdot t
\0&=v_0+gcdot t_{up}rightarrow t_{up}=-dfrac{v_0}{g}
end{aligned}$$

$$begin{aligned}
t=2cdot t_{up}&=2cdotdfrac{v_0}{g}
\&=2cdotdfrac{22.5}{9.81}
\&=boxed{4.59mathrm{~sec}}
end{aligned}$$

$mathrm{bold b)}$ $t=4.59mathrm{~sec}$