Answer:
a) The angle of elevation is 53.1°
b) The speed of the water at the highest point is 15.0 m/s.
c) The magnitude of the acceleration is 9.8 m/s².
d) The water strikes the building at a height of 15.9 m.
e) The speed of the water just before it hits the building is 17.7 m/s
Explanation:
Hi there!
Please, see the attached figure for a better understanding of the problem.
The equation of the position vector of the stream of water at a time "t" is the following:
r = (x0 + v0 · t · cos(α), y0 + v0 · t · sin(α) + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = angle of elevation.
y0 = initial vertical positon.
g = acceleration due to gravity.
We know that at t = 3.00 s the water reaches the building. That means that at t = 3.00 s, the horizontal component of the position vector is equal to 45.0 m. So, using the equation of the horizontal component of the position vector we can obtain the angle α.
x = x0 + v0 · t · cos(α)
Since the origin of the frame of reference is located at the end of the hose, x0 = 0. Then:
x = v0 · t · cos(α)
at t = 3.00 s, x = 45.0 m
45.0 m = 25.0 m/s · 3.00 s · cos(α)
45.0 m / 75.0 m = cos(α)
α = 53.1°
The angle of elevation is 53.1°
b) Let´s see if the water reaches its highest height before it reaches the building. At the highest point, the vertical component of the velocity vector is zero (see v1 in the figure).
The equation of the velocity vector is the following:
v = (v0 · cos(α), v0 · sin(α) + g · t)
We have to find the time at which the y-component of the velocity vector is zero:
vertical component of the velocity vector = vy = v0 · sin(α) + g · t
0 = 25.0 m/s · sin(53.1°) - 9.8 m/s² · t
-25.0 m/s · sin(53.1°) / - 9.8 m/s² = t
t = 2.04 s
If the water reaches the building at t = 3.00 s, then, it reaches its highest point at t = 2.04 s.
Let´s calculate now the horizontal component of the velocity at the highest point which is the same at any point.
vx = v0 · cos(α)
vx = 25.0 m/s · cos(53.1°)
vx = 15.0 m/s
The velocity vector at the highest point of the trajectory is the following:
v = (15.0 m/s, 0)
The magnitude of this vector is 15 m/s. So, the speed of the water at the highest point is 15.0 m/s.
c) The water is only accelerated in the vertical direction by the downward force of gravity (because we neglect air resistance). Then, at the highest point (and at any point) the acceleration vector of the water is:
a = (0, -9.8) m/s²
The magnitude of the acceleration is 9.8 m/s²
d)Let´s find the value of the y-component of the position vector at t = 3.00 s (when the water reaches the building):
y = y0 + v0 · t · sin(α) + 1/2 · g · t² (y0 = 0)
y = 25.0 m/s · 3.00 s · sin(53.1°) - 1/2 · 9,8 m/s² · (3.00 s)²
y = 15.9 m
The water strikes the building at a height of 15.9 m.
e) Using the equation of the velocity vector, let´s calculate the velocity at t = 3.00 s (v2 in the figure):
v2 = (v0 · cos(α), v0 · sin(α) + g · t)
v2x = 15.0 m/s
v2y = v0 · sin(α) + g · t
v2y = 25.0 m/s · sin(53.1°) - 9.8 m/s · 3.00 s
v2y = -9.41 m/s
Then:
v2 = (15.0 m/s, -9.41 m/s)
The magnitude of the velocity vector will be:
[tex]|v| = \sqrt{(15.0 m/s)^{2} + (-9.41 m/s)^{2}} = 17.7 m/s[/tex]
The speed of the water just before it hits the building is 17.7 m/s
