a ball was dropped off a cliff and took 3.59s to fall to the water below. neglecting air resistance how fast in m/s was the ball moving just as it hit th water
v= g•t=9.8•3.59 =35.18 m/s
To answer this question, we can use the equation for free fall motion:
\(v = g \cdot t\)
where:
v = final velocity (speed)
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
t = time
Given that the ball took 3.59 seconds to fall and neglecting air resistance, we can calculate the speed of the ball just as it hit the water.
\(v = 9.8 \, \text{m/s}^2 \cdot 3.59 \, \text{s}\)
Now let's do the calculation:
\(v = 35.282 \, \text{m/s}\)
Therefore, the ball was moving at approximately 35.282 m/s just as it hit the water.