Answer
a) $13.3\;\rm m/s$
b) $48^\circ$
Work Step by Step
Known about the ball:
- $x_i=0\;\rm m$
- $x_f=18\;\rm m$
- $y_i=1\;\rm m$
- $y_f=1\;\rm m$
a)
First of all, we need to find the height of the house which will be the maximum height of the ball so it can pass above the top of the house without hitting it.
From the geometry of the top part of the house, as we see below, we can see that the maximum height of the house is 6 m above the ground.
So, he needs to toss the ball in such a way that makes its height above the top of the house higher than 6 m or barely higher than 6 m.
Thus,
$$y_{max}=6\;\rm m$$
This maximum height should occur just in the middle of the ball's trip which means at
$$x_1=9\;\rm m$$
Now we can find the vertical velocity component by using the kinematic formula of
$$v_{fy}^2-v_{iy}^2=2a_y\Delta y$$
where $\Delta y=5 \;\rm m$ which is the height difference between the point the ball was released and the maximum height of the house, and $a_y=-g$.
We also know that at the highest point the vertical velocity component must be zero.
$$0 -v_{iy}^2=-2g\Delta y$$
$$v_{iy}=\sqrt{2\cdot 9.8\cdot 5}=\bf 9.899\;\rm m/s$$
Now we can find the time for the whole trip
$$y_f=y_i+v_{iy}t-\frac{1}{2}gt^2$$
$$0=0+9.899t-4.9t^2$$
Thus,
$t=0\;\rm s$ (dismissed) or $t=2.02\;\rm s$
Now we can find the horizontal velocity component.
$$v_{ix}=\dfrac{x_f-x_i}{t}=\dfrac{18-0}{2.02}=\bf 8.911\;\rm m$$
Thus, the initial velocity is given by applying the Pythagorean theorem.
$$v_i^2=\sqrt{v_{ix}^2+v_{iy}^2}=\sqrt{8.911^2+9.899^2}$$
$$\boxed{v_i=\color{red}{\bf 13.3}\;\rm m/s}$$
b) We can find the initial angle of the initial velocity by
$$\theta_i=\tan^{-1}\left(\dfrac{v_{iy}}{v_{ix}}\right)=\tan^{-1}\left(\dfrac{9.899}{8.911}\right)$$
$$\boxed{\theta_i=\color{red}{\bf48^\circ}}$$
