Respuesta :
Answer and Solution:
As per the question:
Velocity of water, [tex]v_{o} = 25.0\ m/s[/tex]
Time taken by the water to reach the building, t = 3.00 s
Distance of the building, d = 45.0 m
Now,
(a) For the angle of elevation of the hose, [tex]\alpha[/tex]:
[tex]d = v_{o}tcos(\alpha)[/tex]
[tex]45 = 25\times 3cos(\alpha)[/tex]
[tex]45 = 75cos(\alpha)[/tex]
[tex]\alpha = cos^{- 1}(0.6) = 53.13^{\circ}[/tex]
(b) At the highest point in the trajectory:
Speed = [tex]v_{o}cos\alpha = 25cos53.13^{circ} = 15\ m/s[/tex]
Acceleration, a is due to gravity acting vertically downwards;
a = - g = 9.8[tex]m/s^{2}[/tex]
(c) The height at which the water strikes the building is given by:
H = [tex]v_{o}sin\alpha t - \frac{1}{2}gt^{2}[/tex]
H = [tex]25.0\times 3sin(53.13^{\circ}) - \frac{1}{2}\times 9.8\times 3^{2}[/tex]
H = 15.89 m
Now, the vertical component of the velocity is v' at t = 3.00 s is given by;
v' = [tex]v_{o}sin\alpha - gt[/tex]
v' = [tex]25sin(53.13^{\circ}) - 9.8\times 3 = - 9.4 m/s[/tex]
Now, the resultant velocity from the horizontal and vertical component is given by:
v = [tex]\sqrt{v_{o}^{2} + v'^{2}} = \sqrt{25^{2} + (- 9.4)^{2}} = 26.71\ m/s[/tex]
