The data about the previous problem are missing: I found them online and attached here (see picture). The relevant missing data is just the height of the building, which is 45 meters.
(a) 1.51 s
The motion along the x-direction is uniform (constant speed), so the position along the x-direction is given by
[tex]x(t) = (v_0 cos \theta) t[/tex]
where [tex]v_0[/tex] is the initial speed, [tex]\theta=39.0^{\circ}[/tex] is the angle of launching, t is the time. We can re-write the formula as
[tex]v_0 t = \frac{x(t)}{cos \theta}[/tex] (1)
Along the y-direction the motion is accelerated (free-fall), so the equation for the vertical position is
[tex]y(t)=h-v_0 sin \theta t -\frac{1}{2}gt^2[/tex] (2)
where h = 45.0 m is the height of the building and [tex]g=9.8 m/s^2[/tex] is the acceleration of gravity. Substituting (1) into (2),
[tex]y(t)=h-x(t) tan \theta -\frac{1}{2}gt^2[/tex]
The ball strikes the ground when y(t)=0, so the equation becomes
[tex]0=h-x(t) tan \theta -\frac{1}{2}gt^2[/tex]
And when this happens, the displacement along the x-axis is x = 41.8 m. Solving the equation for t and substituting the numbers, we find the time of flight:
[tex]t=\sqrt{\frac{2(h-xtan \theta)}{g}}=\sqrt{\frac{2(45-41.8 tan 39^{\circ})}{9.8}}=1.51 s[/tex]
(b) 35.6 m/s
We can now easily find the initial speed by looking at the equation for the horizontal displacement:
[tex]x(t) = (v_0 cos \theta) t[/tex]
Re-arranging it,
[tex]v_0=\frac{x(t)}{t cos \theta}[/tex]
And substituting:
x = 41.8 m
t = 1.51 s
[tex]\theta=39.0^{\circ}[/tex]
we find
[tex]v_0 = \frac{41.8}{(1.51)(cos 39.0^{\circ})}=35.6 m/s[/tex]
(c) 46.4 m/s at [tex]53.3^{\circ}[/tex] degrees below the horizontal
The horizontal component of the velocity during the whole motion is constant, so it is
[tex]v_x = v_0 cos \theta = (35.6)(cos 39.0^{\circ})=27.7 m/s[/tex]
The vertical velocity instead follows the equation
[tex]v_y(t) = -v_0 sin \theta -gt[/tex]
where the negative signs means the direction is downward. Substituting t=1.51 s, we find the vertical velocity at the time of impact:
[tex]v_y =-(35.6)(sin 39.0^{\circ})-(9.8)(1.51)=-37.2 m/s[/tex]
Therefore, the speed at the moment of impact is:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(27.7)^2+(-37.2)^2}=46.4 m/s[/tex]
And the direction will be given by
[tex]\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-37.2}{27.7})=-53.3^{\circ}[/tex]
