a) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
[tex]x(t)=v_0 \cos \alpha t[/tex]
[tex]y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2 [/tex]
where
- the horizontal motion is a uniform motion, with constant speed [tex]v_0 \cos \alpha[/tex], where [tex]v_0 = 8.00 m/s[/tex] and [tex]\alpha=20.0^{\circ}[/tex]
- the vertical motion is an uniformly accelerated motion, with constant acceleration [tex]g=9.81 m/s^2[/tex], initial position h (the height of the building) and initial vertical velocity [tex]v_0 \sin \alpha[/tex] (with a negative sign, since it points downwards)
The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
[tex]x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m[/tex]
b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
[tex]0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2 [/tex]
which becomes
[tex]h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m [/tex]
c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
[tex]y(t)=h-10[/tex]
If we substitute this into the equation of y(t), we have
[tex]h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2 [/tex]
[tex] \frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0 [/tex]
[tex]4.9 t^2 +2.74 t-10 =0[/tex]
whose solution is [tex]t=1.18 s[/tex] (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.
