Answer:
A) Parabola; B) 3.83 m; C) 8.84 m; D) 10 m/s
Step-by-step explanation:
A) Shape of water jet
The water jet has the shape of a parabola.
B) Maximum height
Data:
θ = 60°
u = 10 m/s
a = 9.8 m·s⁻²
Calculations:
1. Calculate the horizontal and vertical components of the velocity
[tex]u_{\text{h}} = u \cos \theta = \text{10 m/s} \times \cos 60 ^{\circ} = \text{10 m/s} \times 0.5 = \text{5 m/s}\\u_{\text{v}} = u \sin \theta = \text{10 m/s} \times \sin 60 ^{\circ} = \text{10 m/s} \times 0.866 = \text{8.66 m/s}[/tex]
2. Maximum height
[tex]H = \dfrac{(u_{\text{v}})^{2}}{2a} = \dfrac{8.66^{2}}{2\times 9.8} =\textbf{3.83 m}[/tex]
C) Range
1. Calculate the time of flight
Use the vertical component of velocity to calculate the time to the maximum height of the stream.
[tex]u_{\text{v}} = at\\t = \dfrac{ u_{\text{v}}}{a} = \dfrac{\text{8.66 m$\cdot$s}^{-1}}{\text{9.8 m$\cdot$s}^{-2}}= \text{0.884 s}[/tex]
It will take the same time to reach the ground.
Thus,
Time of flight = 2t = 2 × 0.884 s = 1.77 s
2. Calculate the horizontal distance
s = vt = 5 m·s⁻¹ × 1.77 s = 8.84 m
You should place the drain 8.84 m from the pipe.}
D) Modulus of velocity
The stream of water will hit the drain with the same velocity as when it left the pipe.
Thus, the modulus of the velocity is 10 m/s.
The graph below shows the trajectory of the water stream.
