Help me solve this please

Question

Water from a fire hose is directed toward a building as shown in the figure below. The water leaves the hose at a speed of vi = 40.0 m/s and at an angle of 𝜃i = 39.0° above the horizontal. The base of the hose (at ground level) is a horizontal distance d = 41.0 m away from the building. Find the height h (in m) where the water strikes the building.

Answer 1

The water from the fire hose follows a projectile path. So, we need to use the projectile equations of motion in x and y direction. These are given as

y=u sinθ t -1/2gt^2 ..........(1)

x=u cosθ t ...................(2)

Here x and y are horizontal and vertical distance covered in terms of time t,

Given that

u=40 m/s, θ=39 deg

and g=9.8 m/s^2

Now given that the hose is at the distance of 41 m from the building. So we need to find the time t, taken by water to cover this horizontal distance using Eq. (2). Then we will plug that time in equation (1), to find the vertical distance or height where water will hit the building.

Now, from equation (2), we get

41=40 cos (39)t => t= 1.32 secs

putting it in equation (1), we get

y= 40 sin(39)*1.32 -9.8*(1.32)^2/2= 24.69 m

So the water will hit the building at height 24.69 m above the ground.