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Viefleur [7K]
3 years ago
12

A plant box falls from the windowsill 25.0 m above the sidewalk and hits the cement 3.0 s later. What is the box's velocity when

it hits the ground?
24 m/s
22 m/s
120 m/s
Physics
1 answer:
mamaluj [8]3 years ago
8 0

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.

Substituting for the values in the question we get:

v²=2×9.8m/s²×25m

v²=490m²/s²

v=22.14m/s which can be approximated to 22m/s

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3 years ago
A stone thrown horizontally from a height of 5.72 m hits the ground at a distance of 13.30 m. Calculate the initial speed of the
kakasveta [241]

Answer:

initial velocity=12.31 m/s

Final speed= 16.234 m/s

Explanation:

Given Data

height=5.72 m

distance=13.30 m

To Find

Initial Speed=?

Solution

Use the following equation to determine the time of the stone is falling.  

d = vi ×t   ½ ×9.8 × t²

Where  

d = 5.72m and vi = 0 m/s

so  

5.72 = ½× 9.8 ×t²

t = √(5.72 ÷ 4.9)

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d = v×t

13.30 = v ×1.08

v = 13.30 ÷ 1.08

v=12.31 m/s

To determine stone’s final vertical velocity use the following equation

vf = vi+9.8×t............vi=0 m/s

vf = 9.8×1.08

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