#### Known

The vertical and horizontal motions of the ball are independent. Therefore, in the vertical direction ($y$-axis).

$$
begin{align}
y_{tx{f}}=y_{tx{i}}+v_{tx{y},tx{i}}cdot t+frac{1}{2}a_{tx{y}}cdot t^2
end{align}
$$

#### Calculation

Givens: $y_{tx{i}}=1.25 tx{m}$, $y_{tx{f}}=0 tx{m}$, $v_{tx{y},tx{i}}=0 frac{tx{m}}{tx{s}}$, $g=9.81 frac{tx{m}}{tx{s}^2}$

From (1) we have:

$$
begin{align*}
0 tx{m}=1.25 tx{m}+left(0 frac{tx{m}}{tx{s}}right)t-frac{1}{2}left(9.81 frac{tx{m}}{tx{s}^2}right)t^2, tx{with} a_{tx{y}}=-g\
implies t=sqrt{frac{2(0 tx{m}-1.25 tx{m})}{-9.81 frac{tx{m}}{tx{s}^2}}}=0.505 tx{s}
end{align*}
$$

—

#### Conclusion

The ball falls to the ground in $0.505 tx{s}$

#### Known

The vertical and horizontal motions of the ball are independent. Therefore, in the vertical direction ($y$-axis).

$$
begin{align}
y_{tx{f}}=y_{tx{i}}+v_{tx{y},tx{i}}cdot t+frac{1}{2}a_{tx{y}}cdot t^2
end{align}
$$

#### Calculation

Givens: $y_{tx{i}}=52 tx{cm}=0.52 tx{m}$, $y_{tx{f}}=0 tx{m}$, $v_{tx{y},tx{i}}=0 frac{tx{m}}{tx{s}}$, $g=9.81 frac{tx{m}}{tx{s}^2}$

From (1) we have:

$$
begin{align*}
0 tx{m}=0.52 tx{m}+left(0 frac{tx{m}}{tx{s}}right)t-frac{1}{2}left(9.81 frac{tx{m}}{tx{s}^2}right)t^2, tx{with} a_{tx{y}}=-g\
implies t=sqrt{frac{2(0 tx{m}-0.52 tx{m})}{-9.81 frac{tx{m}}{tx{s}^2}}}=0.33 tx{s}
end{align*}
$$

This is the time it takes for the arrow to travel the vertical ($x$-axis) of $52 tx{cm}$, while traveling a horizontal distance ($x$-axis) of $15 tx{m}$.

—

#### Conclusion

The arrow reaches its target in $0.33 tx{s}$.