Answer :
Final answer:
The water from a water pistol aimed horizontally will drop approximately 0.27 meters while it moves 1.55 meters horizontally. This is found by considering the time taken to move horizontally and applying the kinematics equation for constant acceleration due to gravity.
Explanation:
The question involves calculating how far a stream of water from a water pistol aimed horizontally drops in moving a certain horizontal distance. This is a classic physics problem relating to projectile motion, where gravity is the only force acting on the projectile (in this case, the water stream) after it is released. Since horizontal and vertical motions can be considered independently, we treat the horizontal motion as constant velocity and the vertical motion as constant acceleration due to gravity (9.81 m/s2).
To find the vertical distance the water falls, we use the kinematic equation for constant acceleration:
s = ut + (1/2)at2
Here, s is the vertical displacement, u is the initial vertical velocity (which is 0 since the water is aimed horizontally), a is the acceleration due to gravity, and t is the time in seconds. We solve for t using the horizontal motion:
t = Horizontal Distance / Horizontal Speed
So, t = 1.55 m / 6.55 m/s ≈ 0.2366 s
Plugging this into the vertical motion equation, we get:
s = 0 + (1/2) * 9.81 m/s2 * (0.2366 s)2
s ≈ 0.27 m
Thus, the water drops approximately 0.27 meters in moving 1.55 meters horizontally.
