Question

a stone is thrown vertically upwards with an initial velocity of 48 m /s find (1) the maximum height to Which it rises (2) the total time is taken to return to the earth?​

Answer 1

Given:

• initial velocity of ball (u) = 49 m/s

• final velocity of ball (v) = 0 m/s

To find:

• i) the maximum height at which it rises,

• ii) the total time it takes to return to the surface of the earth.

formula used:

$\\\bigstar\:{\underline{\boxed{\bf{\purple{v^2 - u^2 = 2gh}}}}}$

$\bigstar\:{\underline{\boxed{\bf{\red{v = u + g \times t}}}}}$

solution:

let's

• the time taken is "t"
• and to reach the maximum height "h".

⠀⠀⠀⠀

$\dashrightarrow\sf{v^2 - u^2 = 2gh}$

${\dag}\:{\underline{\bf{putting\:given\:values\:in\:formula,}}}$

$\dashrightarrow\sf (0)^2 - (49)^2 = 2 \times (-9.8) \times h$

$\dashrightarrow\sf 0 - 2401= - 19.6 \times h$

$\dashrightarrow\sf - 2401 = - 19.6 h$

$\dashrightarrow\sf h = \cancel{\dfrac{- 2401}{ - 19.6}}$

$\dashrightarrow{\frak{ \red{h = 160\:m}}}\:\bigstar$

now,

⠀⠀⠀⠀

$\dashrightarrow\sf{v = u + g \times t}$

${\dag}\:{\underline{\bf{\:putting\:given\:values\:in\:formula,}}}$

$\dashrightarrow\sf 0 = 49+ (-9.8) \times t$

$\dashrightarrow\sf 0 = 49- 9.8t$

$\dashrightarrow\sf 9.8t = 49$

$\dashrightarrow\sf t = \cancel{\dfrac{49}{9.8}}$

$\dashrightarrow{\frak{ \pink{t = 5\:sec\:}}}\:\bigstar$

Therefore,

• The total time taken to reach the ground is ( 5 + 5 ) = 10 sec

Answer 2

$\huge\boxed{\bf{\red{ANSWER}}}$

The total time taken to reach the ground is ( 5 + 5 ) = 10 sec

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