Question

An object is dropped from a tower of 180 m height how long does it take to reach the ground take g= 10 m/ssquare

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer}}}}$

From the Question,

• Height of the tower,h = 180m

• Acceleration due to gravity,g = 10m/s²

• Initial Velocity,u = 0 m/s

To find

Time taken to reach the ground

The object is under Free Fall

Thus,

$\large{\sf{h = \frac{1}{2}gt {}^{2} }} \\ \\ \implies \: \boxed{\sf{t = \sqrt{2 \frac{h}{g} } }}$

Putting the values,we get:

$\sf{t = \sqrt{2 \frac{180}{10} } } \\ \\ \rightarrow \: \sf{t = \sqrt{36} } \\ \\ \rightarrow \: \sf{t = \pm \: 6s}$

Since,time can't be negative

$\rightarrow \: \underline{ \boxed {\sf{t = 6s}}}$

Thus,the time taken to reach the ground is 6s

Answer 2

$\huge\underline\mathfrak\red{Answer:-}$

Object takes 6 s to reach the ground.

______________________

$\huge\underline\mathfrak\red{Explanation:-}$

Given :

★ Initial Velocity (u) = 0

★ acceleration due to gravity = 10 m/s²

★ height of tower = 180 m

To find :

★ Time taken by the object to reach the ground.

Solution :

By using second equation of motion,

S = ut + ½at²

But ATQ the equation will be :

h = ut + ½gt²

Putting the given values :

180 = 0(t) + ½(10)(t²)

=> 180 = 5t²

=> t² = 180/5

=> t² = 36

=> t = ±√36

=> t = ±6

Rejecting the negative value, as time can't be Negative.

We get time taken = 6 s

$\huge\boxed{Time\:Taken\:=\:6\:s}$

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Hence, object takes 6 s to reach the ground.