Answer:
2.3s
Explanation:
The initial velocity in the vertical direction:
[tex]v_v = v_0sin(\alpha) = 16.1sin(78^0) = 16.1*0.208 = 3.35 m/s[/tex]
Since the ball is kicked from H1 = 32m to another building at H2 = 13.8m. Vertically speaking it has traveled a distance of 13.8 - 32 = -18.2 m with respect to the initial building
We can use the following equation of motion to solve for the time t it takes to travel
[tex] s = v_vt + gt^2/2[/tex]
where s = -18.2 m is the distance traveled with respect to the first building, g = -9.81 m/s2 is the gravitational acceleration in the opposite direction with the initial velocity
[tex]-18.2 = 3.35t -9.81 t^2/2[/tex]
[tex]4.905t^2 - 3.35t - 18.2 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{3.35\pm \sqrt{(-3.35)^2 - 4*(4.905)*(-18.2)}}{2*(4.905)}[/tex]
[tex]t= \frac{3.35\pm19.19}{9.81}[/tex]
t = 2.3 or t = -1.61
Since t can only be positive we will pick t = 2.3s
