Question

A stone is thrown vertically upward with the velocity of 20m/s.find the maximum height reached by stone and the total time of flight​

Answer 1

Given:-

• Final velocity ,v = 0m/s

• Initial velocity ,u = 20m/s

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

To Find:-

• Maximum height ,h

• Time taken ,t

Solution:-

Firstly we calculate the maximum height attained by the stone.

Using 3rd Equation of Motion

• v² = u² +2ah

v is the final velocity

g is the acceleration due to gravity

u is the initial velocity

h is the height attained by stone

Substitute the value we get

→ 0² = 20² + 2× (-9.8) × h

→ 0 = 400 + (-19.6 ) × h

→ -400 = -19.6×h

→ h = -400/-19.6

→ h = 400/19.6

→ h = 20.40 ≈ 20 m

Therefore, maximum height attained by the stone is 20 metres.

Now, Calculating the time taken

Using 1st Equation of Motion

• v = u +at

Substitute the value we get

→ 0 = 20 + (-9.8) ×t

→ -20 = -9.8×t

→ t = -20/-9.8

→ t = 20/9.8

→ t = 2.04 s

Therefore,. the time taken by the stone to reach maximum height is 2.04 second .

Answer 2

Given :-

Final velocity (V) = 0 m/s [As it stops]

Initial velocity (U) = 20 m/s

Acceleration (A) = 9.8 m/s²

To find :-

• Maximum height
• Time taken

Solution :-

Firstly we will calculate maximum height by using newton third Equation.

$\bf \: v { }^{2} - {u}^{2} = 2ah$

$\sf \: {0}^{2} - {20}^{2} = 2 \times - 9.8 \times h$

$\sf \: 0 - 400 = - 19.6h$

$\sf \: - 400 = - 19.6 \: h$

$\sf \: h = \dfrac{400}{19.6}$

$\sf \: h \: = 20.40 \: m \: \approx \: 20 \: m$

Therefore maximum height travel by stone = 20 M

Now,

Calculating time taken by using newton first Equation

$\bf v = u + at$

$\sf \: 0 = 20 + - 9.8 \times t$

$\sf \: 0 - 20 = - 9.8 \times t$

$\sf - 20 = - 9.8t$

$\sf \: t \: = \dfrac{ - 20}{ - 9.8}$

$\sf \: t \: = 2.04 \: sec$

Therefore, Time taken by stone to travel = 2.04 sec