Question

A body is dropped from the top of a tower. It acquires a velocity of 20 m/s on reaching the ground. Calculate the height of the tower. (Take g = 10 m/s2)

Answer 1

Hey Buddy

Here's The Answer

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Given :-

Final velocity ( v ) = 20 m/s

Initial velocity ( u ) = 0 m/s

Acceleration ( a = g ) = 10 m/s²

We have to find height

We know,

=> v² - u² = 2 × a × s

Here ( s = height ( h ) )

=> v² - u² = 2 × a × h

=> 20² - 0² = 2 × 10 × h

=> 400 - 0 = 20 h

=> 400 = 20 h

=> h = 400/20

=> h = 20 m

So, the height of the tower is 20 m.

Hope It Helps.

Answer 2

Given:-

• Initial velocity(u) = 0m/s
• Final velocity(v) = 20m/s
• Gravitational force (g) = 10m/s²

To find:-

• Height of the tower.

Solution :-

Gravity is cause of an object to fall toward the ground surface at a faster velocity, the longer the object falls.

There is a relation between velocity and gravity, 2as = v² - u², where acceleration(a) = gravity(g) and s = h(height).

✏ From 3rd equation of motion,

$\sf \implies \: 2gh = v {}^{2} - u {}^{2}$

$\sf \implies \: 2 \times 10 \times h = (20) {}^{2} - (0) {}^{2}$

$\sf \implies \: 20h = 400$

$\sf \implies \: h = \dfrac{400}{20}$

$\large \implies \boxed {\boxed {\tt \purple { h = 20 m }}}$

Therefore, height of tower is 20m.