Final answer:
To determine how far the stream of water will land from the base of the cylinder when exiting a hole 5 cm below the top level, apply Torricelli's theorem and projectile motion equations. Calculate the speed of the water using v = √(2gh), and use this with the time it takes for the water to fall the height of the hole to calculate the horizontal distance.
Step-by-step explanation:
The question looks to find the horizontal distance that water would land from the base of the cylinder when it is ejected through a hole on its side. To solve this, we can apply the principles of kinematics and fluid dynamics, particularly Torricelli's theorem for the speed of the water exiting the hole and the projectile motion equations for the distance traveled.
Firstly, the speed of water exiting the hole can be found using Torricelli's Law, which is given by the equation v = √(2gh), where v is the speed of the water, g is the acceleration due to gravity (9.81 m/s²), and h is the height of the water column above the hole, which would be 15 cm or 0.15 m given that the hole is 5 cm below the top level of a 20 cm tall cylinder.
After finding the speed, we use the horizontal projectile motion equation d = vt, where d is the horizontal distance, v is the horizontal velocity (which is the same as the velocity of water exiting the hole for horizontal ejection), and t is the time of flight. Since the water also falls due to gravity, we can use the time it takes to fall 5 cm (0.05 m below the hole) to find t, which can be calculated using the equation t = √(2s/g), where s is the vertical distance fallen.
The final horizontal distance can then be calculated by multiplying the time of flight by the horizontal velocity. However, some additional information or assumptions would be needed to provide a complete solution to this problem.