Question

How long does it take for a brick to reach the ground if dropped from a height of 45m? What will be its velocity just before it reach
the ground?
(A) 2s, 20m/s
(B) 3.645, 35.67m/s
o @ @ E
(C) 38, 30m/s
(D) 5s, 25.10m/s​

Answer 1

Answer:

The answer is a option B

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Answer 2

Question:

How long does it take for a brick to reach the ground if dropped from a height of 45 m? What will be its velocity just before it reaches the ground?

Options:

(A) 2s,20 m/s

(B) 3.645s, 35.67 m/s

(C) 38s, 10 m/s

(D) 5s, 25.10 m/s

(E) 3.03s, 29.7 m/s

Answer:

$\bigstar{\bold{Option\:E=3.03\:s,29.7m/s}}$

Explanation:

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

• Height = Distance travelled (s) = 45 m
• Initial velocity (u) = 0 m/s

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

• Time taken (t)

• Final velocity (v)

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

Time taken:

➻ By the second equation of motion, we know that

   s = ut + 1/2 × a × t²

    where a = acceleration due to gravity = g = 9.8 m/s²

➻ Substituting the datas we get,

    45 = 0 × t + 1/2 × 9.8 × t²

      t² = 45 × 2/9.8

      t² = 90/9.8

      t = √9.18

      t = 3.03 s

➻ Hence time taken = 3.03 s

$\boxed{\bold{Time\:taken=3.03\:s}}$

Final velocity:

➻ By the first equation we know that

    v = u + at

➻ Substituting the datas we get,

   v = 0 + 9.8 × 3.03

   v = 29.7 m/s

➻ Hence the final velocity is 29.7 m/s

$\boxed{\bold{Final\:velocity=29.7\:m/s}}$

Hence options E is  correct.

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

The three equations of motion are:

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