Question

A flowerpot falls from a windowsill 25.0 m above the sidewalk.

A. How fast is the flowerpot moving when it strikes the ground??

B. How much time does a cat on the sidewalk below have to move out of the way before the flowerpot hits the ground?

(Please answer with step by step solutions)

Answer 1

A. The object follows uniformly accelerated motion. We can use the formula,
vf^2 - vo^2 = 2gd
where
vf = final or terminal velocity, m/s
vo = initial velocity, m/s
g = acceleration due to gravity = 9.8 m/s^2

Since the object fell, its initial velocity is zero. Substituting,
vf^2 - 0 = 2*9.8*25
vf^2 = 490
vf = ?


B. We can solve for the time before it hits the ground using this formula:
h = vo*t - (1/2)gt^2

Substituting,
-25 = 0 - (0.5)(9.8)(t^2)
-25 = -4.9t^2
5.102 = t^2
t = ?

Hope this helps~ `u`
B.

Answer 2

IONS