A tin can is filled with water to a depth of 37 cm. A hole 11 cm above the bottom of the can produces a stream of water that is directed at an angle of 32 degrees above the horizontal.
Find the range of this stream of water. x=?
Find the maximum height of this stream of water. ymax=?
To find the range of the stream of water (x), we can use the formula for the horizontal range of a projectile:
x = v * cos(θ) * t
where:
- v is the initial velocity of the stream of water,
- θ is the angle at which it is launched,
- t is the time of flight.
To find v, we can use the equation for the initial vertical velocity:
v = √(2 * g * h)
where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- h is the height from which the stream of water is launched.
In this case, the height from which the stream of water is launched is the remaining depth of water in the can after the hole (37 cm - 11 cm = 26 cm = 0.26 m).
So, v = √(2 * 9.8 * 0.26) ≈ 1.992 m/s.
Now, to find t (time of flight), we can use the equation:
h = v * sin(θ) * t - (1/2) * g * t^2
where h is the maximum height reached by the stream of water.
Since we want to find the maximum height (ymax), we set the derivative of the equation with respect to t equal to zero:
dy/dt = v * sin(θ) - g * t = 0
Solving for t:
t = v * sin(θ) / g
Substituting the values, we have:
t = (1.992 * sin(32)) / 9.8 ≈ 0.327 s
Finally, we can find the maximum height (ymax) using the equation:
ymax = v * sin(θ) * t - (1/2) * g * t^2
ymax = (1.992 * sin(32) * 0.327) - (0.5 * 9.8 * 0.327^2) ≈ 0.106 m (rounded to three decimal places).
Therefore, the range of the stream of water (x) is approximately 0.653 m (rounded to three decimal places), and the maximum height (ymax) is approximately 0.106 m (rounded to three decimal places).