#### Known

The speed is the magnitude of the velocity:

$$
begin{align*}
v=sqrt{v_x^2+v_y^2}
end{align*}
$$

#### Calculation

Givens: From example 4.12: The ball passes over the tree at $t=1.76 tx{s}$, $g=9.81 frac{tx{m}}{tx{s}^2}$, $theta=54.0^circ$, $v_i=13.5 frac{tx{m}}{tx{s}}$

The components of $vec{v}_f$ at $t=1.76 tx{s}$ are given by:

$$
begin{align*}
v_{x,f}&=v_i tx{cos}(theta)=left(13.5 frac{tx{m}}{tx{s}}right) tx{cos}(54.0^circ)\
&=7.94 frac{tx{m}}{tx{s}}\
v_{y,f}&=v_i tx{sin}(theta)-gcdot t=left(13.5 frac{tx{m}}{tx{s}}right) tx{sin}(54.0^circ)-left(9.81 frac{tx{m}}{tx{s}^2}right)(1.76 tx{s})\
&=-6.34 frac{tx{m}}{tx{s}}
end{align*}
$$

Therefore:

$$
begin{align*}
v_f=sqrt{v_{x,f}^2+v_{y,f}^2}=sqrt{left(7.94 frac{tx{m}}{tx{s}}right)^2+left(-6.34 frac{tx{m}}{tx{s}}right)^2}=10.2 frac{tx{m}}{tx{s}}
end{align*}
$$

—

#### Conclusion

The speed of the ball over the tree is $10.2 frac{tx{m}}{tx{s}}$.

#### Known

The vertical and horizontal motions of the ball are independent. With:

$$
begin{align}
&x_f=x_i+left(v_i tx{cos} thetaright)t\
&y_f=y_i+left(v_i tx{sin} thetaright)t+frac{a_ycdot t^2}{2}
end{align}
$$

#### Calculation

Given: $v_i=12 frac{tx{m}}{tx{s}}$, $theta=32^circ$, $t=0.50 tx{s}$, $g=9.81 frac{tx{m}}{tx{s}^2}$, $x_i=0 tx{m}$, $y_i=0 tx{m}$

From (1) and (2):

$$
begin{align*}
x_f&=0 tx{m}+left(12 frac{tx{m}}{tx{s}}cdot tx{cos} (32^circ)right)(0.50 tx{s})\
&=5.09 tx{m}\
y_f&=0 tx{m}+left(12 frac{tx{m}}{tx{s}}cdot tx{sin} (32^circ)right)(0.50 tx{s})-frac{left(9.81 frac{tx{m}}{tx{s}^2}right) (0.50 tx{s})^2}{2}, tx{with} a_y=-g\
&=1.95 tx{m}
end{align*}
$$

—

#### Conclusion

At $0.50 tx{s}$, $x_f=5.09 tx{m}$ and $y_f=1.95 tx{m}$.

#### Known

The components of the vector $vec{v}$ of the ball are given by:

$$
begin{align}
v_{x,f}=v_{x,i}=v_i tx{cos}(theta)hspace{0.5cm}tx{and}hspace{0.5cm}v_{y,f}=v_i tx{sin}(theta)-gcdot t
end{align}
$$

#### Calculation

Given: $v_i=12 frac{tx{m}}{tx{s}}$, $theta=32^circ$, $t=0.50 tx{s}$, $g=9.81 frac{tx{m}}{tx{s}^2}$, $x_i=0 tx{m}$, $y_i=0 tx{m}$

From (1) we have:

$$
begin{align*}
v_{x,f}&=left(12 frac{tx{m}}{tx{s}}right) tx{cos}(32^circ)\
&=10.18 frac{tx{m}}{tx{s}}\
v_{y,f}&=left(12 frac{tx{m}}{tx{s}}right) tx{sin}(32^circ)-left(9.81 frac{tx{m}}{tx{s}^2}right)(0.50 tx{s})\
&=1.45 frac{tx{m}}{tx{s}}
end{align*}
$$

—

#### Conclusion

The $x$ component of velocity is $10.18 frac{tx{m}}{tx{s}}$ and the $y$ component of velocity is $1.45 frac{tx{m}}{tx{s}}$.