Answer
(a) $\alpha = 53.1^{\circ}$
(b) v = 15.0 m/s
$a = 9.80~m/s^2$ directed straight down
(c) The water strikes the building 15.9 meters above the ground.
The speed of the water is 17.7 m/s just before it hits the building.
Work Step by Step
(a) $v_x = \frac{x}{t} = \frac{45.0~m}{3.00~s} = 15.0~m/s$
We can use $v_x$ to find the angle $\alpha$.
$v_x = v_0~cos(\alpha)$
$cos(\alpha) = \frac{v_x}{v_0} = \frac{15.0~m/s}{25.0~m/s}$
$\alpha = cos^{-1}(\frac{15.0~m/s}{25.0~m/s}) = 53.1^{\circ}$
(b) At the highest point:
$v_x = 15.0~m/s$
$v_y = 0$
Therefore, the speed $v$ is 15.0 m/s
$a_x = 0$
$a_y = 9.80~m/s^2$ directed straight down
Therefore, the acceleration is $9.80~m/s^2$ directed straight down.
(c) $v_{0y} = v_0~sin(\alpha) = (25.0~m/s)~sin(53.1^{\circ})$
$v_{0y} = 20.0~m/s$
$y = v_{0y}~t - \frac{1}{2}gt^2$
$y = (20.0~m/s)(3.00~s) - (4.90~m/s^2)(3.00~s)^2$
$y = 15.9~m$
The water strikes the building 15.9 meters above the ground.
We can find the components of the water's velocity.
$v_x = 15.0~m/s$
$v_y = v_{0y} - gt = (20.0~m/s) - (9.80~m/s^2)(3.00~s)$
$v_y = -9.40~m/s$
We can use $v_x$ and $v_y$ to find the magnitude of $v$.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(15.0~m/s)^2+(-9.40~m/s)^2}$
$v = 17.7~m/s$
The speed of the water is 17.7 m/s just before it hits the building.
