Question

A body is dropped freely under gravity from the top of a tower of height 78.4m. Calculate:
(i) the time to reach the ground.​

Answer 1

» To Find :

The Time taken to reach the ground.

» Given :

• Height of the Tower = 78.4 m

» We Know :

Second Equation of Motion :

(Falling under Gravity)

$\sf{\underline{\boxed{h = ut + \dfrac{1}{2}gt^{2}}}}$

Where ,

• h = Height
• t = Time taken
• u = Initial velocity
• g = Acceleration due to gravity

» Concept :

According to the question ,as the body is falling free under Gravity , the initial velocity of the body will be Equal to 0 .

Law : "When a Body starts from rest , it's initial velocity is 0".

So in the Second Equation of motion , we can put the value of initial velocity (u) as 0.

Given Equation :

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

Putting u = 0 ,we get :

$\sf{h = 0 \times t + \dfrac{1}{2}gt^{2}}$

$\sf{h = \dfrac{1}{2}gt^{2}}$

Hence , the new Equation formed for motion , when u = 0 , is :

$\sf{\underline{\boxed{h = \dfrac{1}{2}gt^{2}}}}$

» Solution :

• Height of the Tower = 78.4 m

Using the equation , and Substituting the value in it ,we get :

$\sf{h = \dfrac{1}{2}gt^{2}}$

Taking , $\sf{g = 9.8 ms^{-2}}$

$\sf{\Rightarrow 78.4 = \dfrac{1}{2} \times 9.8t^{2}}$

$\sf{\Rightarrow 78.4 \times 2 = 9.8t^{2}}$

$\sf{\Rightarrow 156.8 = 9.8t^{2}}$

$\sf{\Rightarrow \dfrac{156.8}{9.8} = t^{2}}$

$\sf{\Rightarrow \dfrac{\cancel{156.8}}{\cancel{9.8}} = t^{2}}$

$\sf{\Rightarrow 16 = t^{2}}$

Square Rooting on both the sides ,we get :

$\sf{\Rightarrow \sqrt{16} = \sqrt{t^{2}}}$

$\sf{\Rightarrow 4 s = t}$

Hence ,the time taken by the body to reach the ground is 4 s.

» Additional information :

• First law of motion = v = u + at

• Third law of motion = v² = u² + 2as

• Formula for nth second = s = u + ½(2n - 1)

• Equations on motion , when u = 0.

1. v = at
2. v² = 2as

Answer 2

Answer:

a body is dropped freely under gravity from the top of a tower of height 78.4m calculate time to reach the ground velocity with which it will strike the ground ( acceleration = 10m/s^2 ) t=3.9597=4sec aprox