Math, asked by NoNoNo3, 9 months ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the acceleration due to gravity near Earth's surface, the acceleration due to gravity near the surface of the planet is approximately

A) the same
B) twice as great
C) one-half as great
D) four times as great

Answers

Answered by sarojinipanda02
1

Answer:

option d is correct

mark me as brain list I'll follow u

(#^.^#)(◍•ᴗ•◍)❤乂❤‿❤乂♥(✿ฺ´∀`✿ฺ)ノ

Answered by Anonymous
0

The acceleration due to gravity near the surface of a planet will be one-half as great.

Weight of the sphere= 5kg (GIven)

Time = 3 seconds (GIven)

Fall - 22m (GIven)

One of Kinematics's special cases is the Free-Fall problem. The problem of free fall is a problem in cinematics, where the motion is one-dimensional and is centered in the y-axis.

Considering the relationship - yf - yi = vit + 1/2at²

Then, -22 = 0 + 1/2a × (3)²

-22 = 0 + 1/2a × 9

a =- 4.88m/s²

Approximate of one-half of g will be = -9.8 m/s²

Similar questions