Question

an object is dropped in an unknown planet from height 50m,it reaches the ground in 2s .the acceleration due to gravity in this unknown planet is
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Answer 1

Answer:-

25m/s²

Explanation:-

Let the acceleration due to gravity on that planet be gₚ .

The body is dropped from an certain height, which means that the body is freely falling.

In this case :-

• Value of acceleration due to gravity will

be taken as +ve.

• Initial velocity of the object will be zero.

By using the 2nd equation of motion,we get:-

=> h = ut + 1/2gt²

=> h = ut +1/2gₚt²

=> 50 = 0(t) + 1/2×gₚ×(2)²

=> 50 = 1/2×gₚ×4

=> 50 = 2gₚ

=> gₚ = 50/2

=> gₚ = 25m/s²

Thus, value of acceleration due to gravity in the unknown planet

is 25m/s².

Answer 2

$\huge{\underline{\gray{\sf Given:-}}}$

❏Height = 50m

❏Time taken = 2sec

$\huge{\underline{\gray{\sf Find:-}}}$

❏Accleration due to gravity.

$\huge{\underline{\gray{\sf Solution:-}}}$

we, know that

$\large{\underline{\boxed{\sf s = ut + \dfrac{1}{2}gt^2}}\qquad\Bigg\lgroup\sf Using\:2^{nd}\:eq.\:of\:motion\Bigg\rgroup}$

$\purple{\sf where} \gray{\begin{cases} \sf \red{u = 0m/s} \\ \orange{\sf t = 2sec} \\ \pink{\sf s = 50m}\end{cases}}$

$\dashrightarrow\sf s = ut + \dfrac{1}{2}gt^2$

$\dashrightarrow\sf 50= (0)(2)+ \dfrac{1}{2}g(2)^2$

$\dashrightarrow\sf 50= 0 + \dfrac{1}{2}g \times 4$

$\dashrightarrow\sf 50=\dfrac{4}{2}g$

$\dashrightarrow\sf 50=2g$

$\dashrightarrow\sf \dfrac{50}{2}=g$

$\dashrightarrow\sf 25m/s^2=g$

$\therefore\sf g = 25m/s^2$

Hence, acceleration due to gravity in the unknown planet will be 25m/s²