Question

a freely falling object starts from rest. how long will until it hits the ground if it is dropped 36m above the ground?

Answer 1

Answer:

$\boxed{\sf Time \ taken \ by \ the \ object \ to \ hit \ the \ ground = 2.68 \ s}$

Given:

Initial velocity (u) of the object = 0 m/s (Body starts from rest)

Height above the ground (h) = 36 m

Accelration due to gravity (g) = 10 m/s²

To Find:

Time taken (t) by the object to hit the ground

Explanation:

From equation of motion for object under gravity we have:

$\boxed{ \bold{h = ut - \frac{1}{2} g {t}^{2} }}$

Substituting values of h, u & g in the equation we get:

$\sf \implies 36 = 0(t) - \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \sf \implies 36 = \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \sf \implies 36 = 5 {t}^{2} \\ \\ \sf \implies 5 {t}^{2} = 36 \\ \\ \sf \implies {t}^{2} = \frac{36}{5} \\ \\ \sf \implies {t}^{2} = 7.2 \\ \\ \sf \implies t = \sqrt{7.2} \\ \\ \sf \implies t = 2.68 \: s$