A potted plant rests on a window frame above the ground. The potted plant is off the frame and falls from rest to the grass below 1.93 s later. a) How fast is the pot traveling just before impact with the grass? b) How far did the pot fall?

Maverick Avery

Maverick Avery

Answered question

2022-10-16

A potted plant rests on a window frame above the ground. The potted plant is off the frame and falls from rest to the grass below 1.93 s later.
a) How fast is the pot traveling just before impact with the grass?
b) How far did the pot fall?

Answer & Explanation

lipovicai1w

lipovicai1w

Beginner2022-10-17Added 9 answers

The time taken by a potted plant to reach the ground surface is given as
t=1.93 s
(1)
When the plotted plant freely falls down from the window, the initial velocity of the pot is equal to zero.
u=0 m/s
Use the kinematic equation to determine the velocity of the potted plant just before impact with the grass.
v = u + g t = ( 0   m / s ) + ( 9.8   m / s 2 ) ( 1.93   s ) = 18.914   m / s
(2)
When the potted plant freely falls down from the window, the initial velocity of the pot is equal to zero and the initial velocity has no horizontal component of the velocity. So, it does not move horizontal distance when it freely falls. The potted plant falls vertically downward.
d=0 m

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