Question

A stone is thrown vertically upwards to break a mango from the tree, at a height of 20 m from the ground. If it takes 2 sec to reach the mango, the initial speed of the stone, with which it was thrown upwards is m/s?) _m/s. (take g = 10 )​

Answer 1

Given:-

• Height ,h = 20m

• Time taken ,t = 2 s

• Acceleration due to gravity = 10m/s

• Final velocity ,v = 0m/s (as it hits the Mango)

To Find:-

• Initial velocity ,u

Solution:-

Using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 0² = u² + 2×10 × 20

→ 0 = u² + 20×20

→ 0 = u² + 400

→ u² = 400

→ u = √400

→ u = 20m/s

Therefore, the initial velocity of the stone is 20m/s.

Answer 2

$\huge{\underline{\sf Given:-}}$

⛇Height at which stone is thrown = 20m

⛇Time taken to reach the mango = 2sec

$\huge{\underline{\sf Find:-}}$

⛇Initial speed of the stone at which it was thrown upwards

$\huge{\underline{\sf Solution:-}}$

we, know that

$\huge{\underline{\boxed{\sf v^2 - u^2 = 2gs}}} \qquad \bigg\lgroup{ \sf {3}^{rd} \: eq. \: of \: motion} \bigg\rgroup$

where,

• Final Velocity, v = 0m/s
• Here, acceleration due to gravity, g = 10m/s²
• Height, s = 20m

So,

$\dashrightarrow\sf v^2 - u^2 = 2gs$

$\dashrightarrow\sf 0^2 - u^2 = 2(10)(20)$

$\dashrightarrow\sf -u^2 = 2(200)$

$\dashrightarrow\sf -u^2 = 400$

$\dashrightarrow\sf -u = \sqrt{400}$

$\dashrightarrow\sf -u = 20$

$\dashrightarrow\sf 20 + u = 0$

$\dashrightarrow\sf u = - 20m/s$

Here, Speed can't be in negative.

Hence,

• Initial speed, u = 20m/s