You’re 6.0 m from one wall of the house seen in FIGURE P4.55. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground.

a. What minimum speed will allow the ball to clear the roof?

b. At what angle should you toss the ball?

Given the diagram is as follows:

The initial vertical height of the ball is $1\text{m}$, the distance of the thrower is $6.0\text{m}$, the inclination angle of the roof of the house is$45°$.

The top of the house is at the middle from the both ends of the house in the ground.

The height $\left(h\right)$of the top of the house, as can be seen from the diagram, is given by

$\begin{array}{rcl}h& =& 3\text{m}\times \tan 45°\\ & =& 3\text{m}\end{array}$

The total height $\left(h_H\right)$of the house from the ground is given by

role="math" $\begin{array}{rcl}{h}_H& =& 3\text{m}+3\text{m}\\ & =& 6\text{m}\end{array}$

The vertical height $\left(\begin{array}{rcl}& & h_b\end{array}\right)$travelled by the ball is given by

$\begin{array}{rcl}{h}_b& =& 6\text{m}-1\text{m}\\ & =& 5\text{m}\end{array}$

When the ball reaches the maximum height, its final velocity is zero.

The formula to calculate the vertical component of the final speed of the ball is given by

${v_{fy}}^2={v_{if}}^2-2g{h}_b..........................\left(1\right)$

Here, $v_{fi}$is the vertical component of the final speed, $v_{iy}$is the vertical component of the initial speed and $g$is the acceleration due to gravity.

Substitute $0$for $v_{fy}$, $9.80{\text{m/s}}^2$for $g$and $5\text{m}$for $h_b$into equation (1) and solve to calculate the vertical component of the final speed.

$\begin{array}{rcl}0& =& {v_{iy}}^2-2\times 9.80{\text{m/s}}^2\times 5\text{m}\\ {v_{iy}}^2& =& 2\times 9.80{\text{m/s}}^2\times 5\text{m}\\ v_{iy}& =& \sqrt{2\times 9.80{\text{m/s}}^2\times 5\text{m}}\\ & \approx & 9.9\text{m/s}\end{array}$

The formula to calculate the final vertical component of the ball is given by

$v_{fy}=v_{iy}-gt..............\left(2\right)$

Here, $t$is the time taken by the ball to reach the top of the house.

Substitute $0$for $v_{fy}$, $9.9\text{m/s}$for $v_{iy}$and $9.80{\text{m/s}}^2$for $g$into equation (2) and solve to calculate the required time.

$\begin{array}{rcl}0& =& 9.9\text{m/s}-9.8{\text{m/s}}^2\times t\\ t& =& \frac{9.9\text{m/s}}{9.8{\text{m/s}}^2}\\ & \approx & 1.01\text{s}\end{array}$

The formula to calculate the horizontal distance travelled by the ball when it reaches the maximum height is given by

role="math" $d=v_{ix}t....................\left(3\right)$

Here, $d,v_{ix}$are the horizontal distance travelled by the ball and the horizontal component of its initial speed respectively.

Substitute $9\text{m}$for $d$and $1.01\text{s}$for $t$into equation (3) and solve to calculate the horizontal component of the initial speed of the ball.

$\begin{array}{rcl}9\text{m}& =& v_{ix}\times 1.01\text{s}\\ v_{ix}& =& \frac{9\text{m}}{1.01\text{s}}\\ & \approx & 8.91\text{m/s}\end{array}$

The formula to calculate the initial speed $\left(v_i\right)$of the ball is given by

role="math" localid="1647577931949" ${v_i}^2={v_{ix}}^2+{v_{iy}}^2...............\left(4\right)$

Substitute $8.91\text{m/s}$for $v_{ix}$and $9.9\text{m/s}$for $v_{iy}$into equation (4) and solve to calculate the initial speed of the ball.

$\begin{array}{rcl}{v_i}^2& =& \left(8.{91}^2+9.9^2\right){\text{m}}^2{\text{/s}}^2\\ v_i& =& \sqrt{\left(8.{91}^2+9.9^2\right){\text{m}}^2{\text{/s}}^2}\\ & \approx & 13\text{m/s}\end{array}$

The formula to calculate the projection angle $\left(\theta \right)$of the ball is given by

role="math" $\tan \theta =\frac{v_{iy}}{v_{ix}}...............\left(5\right)$

Substitute the values of the parameters into equation (5) and solve to calculate the required projection angle.

$\begin{array}{rcl}\tan \theta & =& \frac{9.9\text{m/s}}{8.91\text{m/s}}\\ \theta & =& {\tan }^{-1}\left(\frac{9.9}{8.91}\right)\\ & \approx & 48°\end{array}$