Question

a stone is dropped from the top of a building how long does it take to fall 19.6m.
how fast does it move the end of this fall?
what is the acceleration after 1 second and 2 seconds? g=10m/s

Answer 1

h = 19.6 m
u = 0
h = u t + 1/2 g t * t
19.6   = 0 + 1/2 * 9.8 * t * t            as  g = 9.8 m/s/s

t = 2 sec.    it takes 2 sec.

acceleration is always the same : g = 9.8 m/sec

velocity / speed at t = 2 sec is  =  u + g  t  = 0 + 9.8 * 2 sec = 19.6 m/sec

Answer 2

Explanation:

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

• it take to fall 19.6m.

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

• How does it take to fall 19.6m

• How fast does it moves at a end of this fall.

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

Total time taken for stone to fall, :

$\boxed{\sf \: g = 9.8 \: {ms}^{2} }$

a) How does it take to fall 19.6m. :

$\sf \leadsto \: 19.6 = 0(t) + \frac{1}{2} (g) {n}^{2} \\ \\ \sf \leadsto \: {n}^{2} = \frac{ \cancel{19.6} \times 2}{ \cancel{9.8} }\\ \\ \sf \leadsto \: {n}^{2} = 2 \times 2 \\ \\ \sf \leadsto \: {n}^{2} = 4 \\ \\ \sf \leadsto \: n = \sqrt{4} \\ \\ \sf \leadsto \: n \: = 2$

Therefore, displacement in last, that is the 2nd second,u

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b) How fast does it moves at a end of this fall.

$\boxed{ \sf \: {v}^{2} = {u}^{2} + 2as }$

Substitute all values :

$\sf \leadsto \: {v}^{2} = 0 + 2 \times 9.8 \times 19.6 \\ \\ \sf \leadsto \: {v}^{2} \: = 384.16 \\ \\ \sf \leadsto \: v \: = \sqrt{384.16} \\ \\ \\ \sf \leadsto \: v = 19.6 m/s^2$