Question

A stone is thrown in a virtically opposite direction with a velocity of 5m/s . If the acceleration of the stone during it's motion is 10 m/s in the downward direction, what will be the hight attained by the stone and how much time will it take to reach there

Answer 1

Given :

➨ A stone is thrown vertically upward at a velocity of 5m/s.

➨ Acc. due to gravity = 10m/s²

To Find :

➾ Max. height attained by ball.

➾ Time taken by ball to reach at there.

Concept :

➜ Since, acceleration due to gravity has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.

➜ For a body thrown vertically upward, g is taken negative.

Calculation :

✭ Maximum height :

⇒ v² - u² = 2(-g)H

⇒ (0)² - (5)² = -20H

⇒ 25 = 20H

⇒ H = 25/20

⇒ H = 1.25m

✭ Time :

⇒ v = u - gt

⇒ 0 = 5 - 10t

⇒ 10t = 5

⇒ t = 5/10

⇒ t = 0.5s

Answer 2

$\bigstar$ $\sf{\red{\underline{\underline{\purple{Answer\::\:1.25\:meter\:and\:0.5\:second}}}}}$

$\bigstar$ $\sf{\red{\underline{\underline{\purple{CONCEPTS:-}}}}}$

☞ when an object aceelerated with constant acceleration, then Equation of Motion is

• $\tt{v\:=\:u\:+\:at}$
• $\tt{s\:=\:ut\:+\:{\dfrac{1}{2}}at^2}$
• $\tt{v^2\:-\:u^2\:=\:2as}$

$\bigstar$ $\sf{\red{\underline{\underline{\purple{To\:Find:-}}}}}$

1. The height attained by the stone .
2. How much time will it take to reach there .

$\bigstar$ $\sf{\red{\underline{\underline{\purple{SOLUTION:-}}}}}$

☞ GIVEN:-

• initial Velocity (u) = 5 m/s
• final velocity (v) = 0

1. since , “u” is upward and “a” is downward, it is a retarded motion .

• acceleration (a) = -10 m/s²

☞ (1) Using the relation, “ v² - u² = 2as ” we have

$\tt{s\:=\:{\dfrac{v^2\:-\:u^2}{2a}}}$

$\tt{\implies\:s\:=\:{\dfrac{0^2\:-\:5^2}{2\times(-10)}}}$

$\tt{\implies\:s\:=\:1.25\:m}$

☞ (2) Using the relation, “ v = u + at ” we have

$\tt{\implies\:0\:=\:5\:+\:(-10)t}$

$\tt{\implies\:t\:=\:{\dfrac{5}{10}}}$

$\tt{\implies\:t\:=\:0.5\:s}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:1.25\:meter\:\:and\:\:0.5\:second\:}}}}$