A flowerpot falls from a window sill 37.6 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
Please show me the formula and steps i would use to solve this, Thanks!
v = a t = 9.81 t
d = (1/2) a t^2 = 4.905 t^2 = 37.6
so
t^2 = 7.67
t = 2.77 s
v = 9.81* 2.77 = 27.2 m/s
that's what i did but the computer is telling me it's wrong
thats wrong mate
To find the velocity of the flowerpot when it strikes the ground, you can use the equation of motion known as the third equation of motion, which relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d):
vf^2 = vi^2 + 2ad
In this case, since the flowerpot is falling vertically downward due to gravity, the initial velocity (vi) is 0 m/s. The acceleration (a) is the acceleration due to gravity, which is given as 9.81 m/s^2. The displacement (d) is the distance the flowerpot falls, which is 37.6 m. Plugging in these values, we can solve for the final velocity (vf).
Replacing the variables in the formula:
vf^2 = 0^2 + 2 * 9.81 * 37.6
Simplifying:
vf^2 = 2 * 9.81 * 37.6
vf^2 = 2 * 365.6556
vf^2 = 731.3112
Taking the square root of both sides:
vf = √731.3112
vf ≈ 27.03 m/s
Therefore, the velocity of the flowerpot when it strikes the ground is approximately 27.03 m/s.