Water flows out horizontalmy at the end of a pipe at a height of 52 cm from the floor.If the horizontal distance before it hits the floor is 100 cm,calculate the velocity of the water at the instant it leaves the pipes?
it takes 0.326 seconds for the water to fall 52 cm.
So, the horizontal speed is 100cm/.326s = 306.9 cm/s
To calculate the velocity of the water at the instant it leaves the pipe, we can make use of basic principles of projectile motion.
First, we need to determine the time it takes for the water to reach the floor. We can do this by considering the vertical motion of the water.
Given that the water falls from a height of 52 cm and the vertical distance it travels is 0 cm (since it reaches the floor), we can use the equation of motion:
h = ut + (1/2)gt^2
where h is the vertical distance (52 cm), u is the initial vertical velocity (0 cm/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken.
Converting the height and acceleration due to gravity to meters, we have:
0.52 m = 0t + (1/2)(-9.8 m/s^2)t^2
0.52 = (-4.9)t^2
Solving for t by rearranging the equation, we get:
t^2 = 0.52 / (-4.9)
t^2 = -0.10612
Since time cannot be negative, we discard the negative solution. The positive value gives us the time taken for the water to reach the floor.
t = √(0.10612)
t ≈ 0.33 s
Now, we can find the horizontal velocity of the water using the horizontal distance traveled before hitting the floor (100 cm) and the time taken (0.33 s).
Horizontal velocity (v) = distance / time
v = 100 cm / 0.33 s
Converting cm to meters and performing the calculation, we find:
v ≈ 303.03 cm/s ≈ 3.03 m/s
Therefore, the velocity of the water at the instant it leaves the pipe is approximately 3.03 m/s.