Question

A stone is thrown vertically upwards with an initial velocity of 40 m/s. Taking g = 10 m/s,
find the maximum height reached by the stone.​

Answer 1

Answer:

Given:

Initial velocity of stone = 40 m/s

Value of g = 10 m/s²

To find

Max height reached

Assumptions:

Value of g doesn't change along the course of trajectory considering height much lesser than radius of Earth

Calculation:

∴ v² = u² - 2gh

=> 0² = 40² - (2 × 10 × h)

=> 1600 = 20h

=> h = 80 metres.

Sign of g is taken negative as the direction of g vector is opposite to the direction of velocity and displacement vector.

So final answer:

$\boxed{ \large{height = 80 \: metres}}$

Answer 2

Given :

• Initial Velocity, u = 40 m/s
• Final Velocity, v = 0 m/s
• g = - 10 m/s

To Find :

• The maximum height reached by the stone (s)

Solution :

Here, we can use the third equation of motion and calculate the height, s reached by the stone with an initial velocity of 40 m/s.

Third Equation Of Motion :

$\bold{\large{\boxed{\mathtt{v^2\:=\:u^2\:+\:2gs}}}}$

Block in the available data,

➟ $\mathtt{0^2\:=\:40^2\:+\:2\:\times\:-10\:\times\:s}$

➟ $\mathtt{0\:=\:1600\:+\:(-20)\:s}$

➟ $\mathtt{-1600\:=\:-20\:s}$

➟ $\mathtt{\dfrac{-1600}{-20}\:=\:s}$

➟ $\mathtt{\dfrac{1600}{20}\:=\:s}$

➟ $\mathtt{s=80\:m}$

$\bold{\sf{\therefore{\underline{Maximum\:height\:reached\:by\:the\:stone\:is\:80\:metre}}}}$