## Fundamentals of Physics Extended (10th Edition)

$$v^{2}_{0}=v^{2}_{0y}+v^{2}_{0x}\Rightarrow v_{0}=\sqrt {v^{2}_{0y}+v^{2}_{0x}}\approx 26\dfrac {m}{s}$$
Let's play it backwards and assume ball is launched from the top of building at point A with an angle of $\theta$ and it reaches the ground after $t=4s$ and the height of building is $h=20m$. So let's write the equation showing height of ball respect to ground at any given time. $v$ is the velocity of the ball at point A $H=h+vt\sin \theta -g\dfrac {t^{2}}{2}$ When it reaches the ground $H=0$ so $0=h+vt\sin \theta -g\dfrac {t^{2}}{2}\Rightarrow 0=20m+v\times 4s\times\sin {60}-\dfrac {9.8\dfrac {m}{s^{2}}}{2}\left( 4s\right) ^{2}$ İf we solve the equation we get $v=16,86\dfrac {m}{s}$ So let's find horizontal component of velocity of the ball which doesn't change because there is no horizontal force acting on the ball $v_{x}=v\cos \theta =16,86\cos 60^{0}=8.43\dfrac {m}{s}$ let's find the vertical component of its velocity at point A $v_{y}=v\sin \theta =16,86m\times\sin {60}=14,6\dfrac {m}{s}$. So the verical component of velocity at point B will be $v_{By}=v_{y}-gt=14,6\dfrac {m}{s}-9,8\dfrac {m}{s^{2}}\times 4s=-24,6\dfrac {m}{s}$ we played the movement backwards so the original initial vertical component of velocity of the ball will be: $v_{oy}=-v_{By}=24,6\dfrac {m}{s}$ Since $v_{0x}=v_{x}=8,43\dfrac {m}{s}$ So the total velocity of the ball at point B (initial velocity) will be : $v^{2}_{0}=v^{2}_{0y}+v^{2}_{0x}\Rightarrow v_{0}=\sqrt {v^{2}_{0y}+v^{2}_{0x}}\approx 26\dfrac {m}{s}$