Firemen are shooting a stream of water at a burning buildingusing a high-pressure hose that shoots out the water with a speedof 25.0 m/s. as it leaves the end of the hose. once it leaves thehose, the water moves in projectile motion. the firemen adjust theangle of elevation α of the hose until the water takes3.00 s to reach the building 45.0 m away. you can ignore airresistance; assume that the end of the hose is at grounglevel.
a) find the angle of elevation α
b) find the speed and acceleration of the water at the highestpoint in its trajectory
c) how high above the ground does the water strike thebuilding, and how fast is it moving just before it hits thebuilding?

a) angle θ = 53.1º, b)    V =  15 i ^ m/s,   a = - 9.8 j ^ m/s² c)   Y = 15.9 m

Explanation:

a)   In the problem they indicate that we treat the drops of water like projectiles, as they give us the time it takes to travel distance we can find the horizontal speed

         Vox = X / t

          Vox = 45.0 / 3.00

          Vox = 15 m / s

They also indicate the speed of water 25m / s, we calculate the angle

         Vox = Vo cos θ

         Vo cos θ = 15

         Cos θ = 15 / Vo

         θ = cos⁻¹ (15 / vo) = cos⁻¹ (15/25)

         θ = 53.1º

This is the elevation angle.

b. At the highest point the vertical speed must be zero (Vy = 0), the constant horizontal speed since there is no acceleration on this axis and the only acceleration acting on the body is the acceleration of gravity

         V = Vox i ^ = 15 i ^ m / s

         a = -g j ^ = - 9.8 j ^ m /s²

Where the negative sign indicates that the acceleration is towards the center of the earth

c. The water takes to reach the building 3 s let's calculate the height

         Y = Voy t - ½ g t²

         Voy = Vo sin θ

         Voy = 25 sin 53.1

         Voy = 20 m / s

         Y = 20 3 - ½ 9.8 3²

         Y = 15.9 m