Question

An object is thrown vertically upward in the sky, it returns to the ground after 6 s. Calculate
(i) the initial velocity of the object
(ii) the maximum height it reaches.

Answer 1

Answer:

$\bigstar{\bold{Initial\:velocity=29.4\:m/s}}$

$\bigstar{\bold{Maximum\:height=44.1\:m}}$

Explanation:

$\Large{\underline{\underline{\sf{Given:}}}}$

• Time taken(t) = 6 s
• Acceleration (a) = Acceleration due to gravity (g) = 9.8 m/s²
• Final velocity (v) = 0 m/s

$\Large{\underline{\underline{\sf{To\:Find:}}}}$

• Initial velocity of the object (u)
• Maximum height it reaches (s)

$\Large{\underline{\underline{\sf{Solution:}}}}$

Initial velocity:

➣ We have to find initial velocity (u) of the object.

➣ By the first equation of motion we know that

    v = u + at

➣ Here the time taken for the total journey is 6 s

➣ Hence time taken for upward journey is 6/2 = 3 s

➣ Substituting the datas we get,

    0 = u + -9.8 × 3

➣ Here g is taken as negative since the motion of the object is in the direction opposite to that of accleration due to gravity.

➣ Solving it we get,

    0 = u - 29.4

    u = 29.4 m/s

➣ Hence the initial velocity of the body is 29.4 m/s

   $\boxed{\bold{Initial\:velocity=29.4\:m/s}}$

Maximum height:

➛ By the second equation of motion we know that,

  s = ut + 1/2 × a × t²

➛ Substituting the datas,

   s = 29.4 × 3 + 1/2 × -9.8 × 3 × 3

   s = 88.2 + -88.2/2

   s = 88.2 - 44.1

   s = 44.1 m

➛ Hence the maximum height reached by the object is  44.1 m

    $\boxed{\bold{Maximum\:height=44.1\:m}}$

$\Large{\underline{\underline{\sf{Notes:}}}}$

→ The three equations of motion are :

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as