Question

A stone is release from the top of tower of height 50m.calculate the velocity just before touching the ground

Answer 1

Answer:

Explanation:

Height = S = 50m

g = acceletration due to gravity on Earth = $9.81 m/s^2$

Now, we will use the equation

$2gS = v^2 - u^2$

Here u = 0 = intital velocity of the stone when released from the tower

Therefore,

$v^2 = 2*9.81*50$

= 981

$v = \sqrt{981} m/s$

v = 32.32 m/s

Therefore, the stone will have attained the speed of about 32.32m/s right before it hits the ground.

Here we have made the regular assumptions such as neglecting air resistance.