Question

A stone is released from the top of a tower of height 20m. Calculate the velocity with which the stone hits the ground. Take the value of 'g' as 10ms.​

Answer 1

Explanation:

$initial \: velocity \: of \: stone \: (u) = 0 \: m {s}^{ - 1} \\ \\ acceleration \: due \: to \: gravity(g) = 10 {ms}^{ - 2} \\ \\ height \: (h) = 20m \\ \\ by \: using \: 3rd \: eq. \: ofmotion \\ \\ 2gh = {v}^{2} - {u}^{2} \\ \\ 2 \times 10 \times 20 = {v}^{2} - 0 \\ \\ {v}^{2} = 400 \\ \\ v = \sqrt{400} \\ \\ v = 20m {s}^{ - 1}$