Question

a body is thrown upward with velocity 30 metre per second from the top of a tower which reaches to the ground after 20 second find height of the tower​

Answer 1

Given:-

• Initial velocity,u = 30 m/s

• Time taken,t = 20 seconds

• Acceleration due to gravity,g = -9.8 m/s²

• Final velocity,v = 0 m/s

To be calculated:-

Calculate the height of the tower?

Formula used:-

v² = u² + 2gh

Solution:-

From the third equation of motion for freely falling body

v² = u² + 2gh

★ Substituting the values in the above formula,we get:

⇒ ( 0 )² = ( 30 )² + 2 × -9.8 × h

⇒ 0 = 900 + ( - 19.6 ) × h

⇒ 0 = 900 - 19.6 h

⇒ 19.6 h = 900

⇒ h = 900/19.6

⇒ h = 9000/196

⇒ h = 45.91 m

Thus,the height of the tower is 45.91 m.

Answer 2

$\bigstar\sf\pink{ANSWER}\bigstar$

GIVEN:-

$\rm\red{Initial\:Velocity(u)=30m/s^{-1}}$

$\rm\blue{Final\:Velocity(v)=0m/s^{-1}}$

$\rm\pink{Time=20 second}$

$\bigstar\rm\blue{HOW\:TO\:SOLVE}\bigstar$

• if we have to find the height of the tower first we have to find the Acceleration from which the body is thrown,then we will use second equation of motion to find the heigjt of the tower.

$\bigstar\rm\blue{FORMULA\:USED}\bigstar$

$huge\boxed\sf\blue{a=\dfrac{v-u}{t}}$

$\huge\boxed\rm\red{S=ut+\dfrac{1}{2}×a×(t)^{2}}$

Now, To find acceleration

$\implies\sf\red{a=\dfrac{0-30}{20}}$

$\implies\sf\blue{a=\dfrac{-30}{20}}$

$\implies\sf\pink{a= 1.5m/s^{-2}}$

Now,To find height of the tower

$\implies\sf\red{S=ut+\dfrac{1}{2}×a×(t)^{2}}$

$\implies\sf\blue{S=30×20+\dfrac{1}{\cancel{2}}×1.5×\cancel{400}}$

$\implies\sf\pink{S=600+300}$

$\implies\sf\red{S=900m}$