A gardener holds a hose 0.75m above the ground such that the water shoot out horizontally and hits the ground at a point 20m away.What is the speed with which the water leaves the hose?
time in air:
hf=hi-1/2 g t^2
0=.75-4.9 t^2 solve for t
then, in the horizontal
20=vi*t solve for vi
To find the speed with which the water leaves the hose, we can use the equations of motion in projectile motion. The horizontal motion of the water is independent of the vertical motion, so we can analyze them separately.
First, let's consider the horizontal motion. The water travels a distance of 20m horizontally, and this motion is unaffected by gravity. The formula for horizontal motion is:
distance = speed * time
Since the water shoots out horizontally, the time taken for it to travel 20m is the same as the time taken for an object to fall 0.75m vertically (because both motions happen simultaneously). We can use the formula for vertical motion to find the time:
distance = (1/2) * acceleration * time²
Plugging in the values for distance (0.75m) and acceleration due to gravity (-9.8 m/s²), we can solve for time:
0.75 = (1/2) * -9.8 * time²
time² = -0.75 / ((1/2) * -9.8)
time = √(0.75 / (0.5 * 9.8))
Now that we have the time, we can substitute it back into the horizontal distance equation to find the speed:
20 = speed * time
speed = 20 / time
Calculating this expression will give us the speed with which the water leaves the hose.