A hose lying on the ground shoots a stream of water upward at an angle of 60 * to the horizontal with a velocity of 16m/s. The height at
Dear Karsang, First we need to get the horizontal and vertical components of the water's velocity. We do this using trigonometry, imagining that the velocity is the hypotenuse of a right-angled triangle and the horizontal and vertical components are the horizontal and vertical sides.
Initial Velocity, Vo =16 m/s
Horizontal Component, Vhor =Vcos(theta)
= 16 × cos 60* = 8 m/s
Vertical Component, Vver = V sin(theta)
= 16 × sin 60* = 13.86 m/s
The next task is to find the time it takes for the water to reach the wall. To do this we can use the horizontal velocity and the distance to the wall.
t = d/v = 8/8 =1 sec.
Now, we cam use the equation
s= ut + 1/2at^2
Where:
s : is the displacement of the body (basically how high up the water hits the wall)
u=is the initial velocity of the body (the vertical component in this case, 13.86 m/s)
a :is the acceleration on the body = - 9.8m/s^2
t : time taken which is 1 sec.
You can put all the values in the equation you will get the result
All the best
Feel free to ask for any query
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