A flowerpot falls from a window sill 28.4 m
above the sidewalk.
What is the velocity of the flowerpot when
it strikes the ground? The acceleration of
gravity is 9.81 m/s2 .
Answer in units of m/s
Initial velocity, (u), is given, so is acceleration, (g), so is the distance (s), 28.4m.
Use v^2=u^2+2as and solve for v.
23.4
To find the velocity of the flowerpot when it strikes the ground, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (which we want to find)
- u is the initial velocity (which is 0 since the flowerpot is initially at rest)
- a is the acceleration due to gravity, which is -9.81 m/s^2 since it acts in the opposite direction of motion (negative due to direction)
- s is the distance traveled by the flowerpot, which is 28.4 m
Plugging in the values, the equation becomes:
v^2 = 0 + 2*(-9.81)*28.4
Simplifying,
v^2 = -559.81
Taking the square root of both sides,
v = √(-559.81)
However, the result is a complex number since the square root of a negative number is an imaginary number. In this case, it is not physically meaningful since we are dealing with real-world motion.
Therefore, the velocity of the flowerpot when it strikes the ground is not a real value, and we cannot provide an answer in units of m/s.