Question

A stone is released from the top of a tower 125 m high. Then find:
(1) the time with which the stone strikes the ground.
(2) the final velocity of the stone. (Take g = 10 m/s2)

Answer 1

Answer:

(1) 5 seconds

(2) 50 m/s

Explanation:

Given,

• Height of the stone, h = 125 m

• initial velocity of the object = 0 m/s

(1) according to equation of motion,

$h\ =\ \dfrac{1}{2}.g.t^2$

$=>\ 125\ m\ =\ \dfrac{1}{2}\times 10\times t^2$

$=>\ t^2\ =\ \dfrac{250}{10}$

$=>\ t^2\ =\ 25$

$=>\ t\ =\ 5$

Hence, time taken by the stone to reach ground 5 seconds.

(2) the final velocity of the stone can be given by using equation of motion,

$v^2\ =\ u^2\ +\ 2gs$

       $=\ 0\ +\ 2\times 10\times 125$

       = 2500

$=>\ v\ =\ 50\ m/s$

Hence, the final velocity of the stone is 50m/s.