Question

An object is allowed to fall freely near the surface of a planet. The object falls 54 m in the first 3.0 seconds after its released. The acceleration due to gravity on that planet is?

a) 1.1 × 10² m/s²
b) 27 m/s²
c) 12 m/s²
d) 6.0 m/s²

PLEASE ANSWER AND EXPLAIN!

Answer 1

Given, distance, s = 54 m

time, t = 3 sec

and initial velocity, u = 0 m/s

acceleration, a = ?

s = ut + 1/2at²

54 = 0(3) + 1/2(a)(3²)

54 = 9/2a

a = 54x2/9

a = 12 m/s²

plz mark it as brainliest

b..bye

#ANSWERWITHQUALITY

#BAL

Answer 2

The value of the acceleration due to gravity on that planet is equal to 12 ms⁻². Hence, the correct option is option (c) 12 ms⁻² .

Given:

Distance up to which the object falls = 54 m

The time taken to fall 54 m = 3 seconds

To Find:

The acceleration due to gravity on the planet.

Solution:

Let the acceleration due to gravity on the planet be 'a'.

→ We would assume the downward direction of acceleration to be positive.

→ The initial velocity of the object (u) = 0

→ The distance up to which the object falls (S) = 54 m

→ The total time taken (T) = 3 seconds

→ Now using the 2nd equation of motion:

                $Second\:equation\:of\:motion:S=ut+\frac{1}{2} at^{2}$

                                 $\boldsymbol{\therefore S=ut+\frac{1}{2} at^{2} }$

                                 $\therefore 54=(0)\times(3)+\frac{1}{2}\times a\times(3)^{2} \\\\\therefore 54=a\times\frac{9}{2} \\\\\therefore a=\frac{(54\times2)}{9} \\\\\therefore a=6\times2\\\\\therefore a=12\:ms^{-2}$

The value of 'a' comes out to be equal to 12 ms⁻².

Therefore the value of the acceleration due to gravity on that planet is equal to 12 ms⁻². Hence, the correct option is option (c) 12 ms⁻².

#SPJ2