Question

A stone thrown vertically upwards with a speed of 5m/s.How much height the stone goes before back to the earth?​

Answer 1

Answer:

Explanation:

Initial Velocity=5m/s

Final velocity= 0

Acceleration=-9.8m/s^2

So height will be,

2as= v^2-u^2

2*-9.8*s=0-25

-19.6s=-25

s = 25/19.6

1.27m

Answer 2

Answer:

Height = 1.25m. Therefore, the stone goes 1.25m high before back to the earth.

Explanation:

Given, initial velocity,

$u = {5ms}^{ - 1}$

Final velocity,

$v = 0$

Since, u is upward & a is downward, it is a retarded motion.

$a = { - 10ms}^{ - 2}$

Height attained by stone, s=?

Time take to attain height, t=?

(i) Using the relation,

${v}^{2} - {u}^{2} = 2as$

we have,

$s = {v}^{2} - {u}^{2} \: or \: 2a \\ = {(0)}^{2} - {(5)}^{2} \: or \: 2 \times ( - 10) = 1.25m$

(ii) Using the relation,

$v = u + at$

$0 = 5 + ( - 10)t \: \: or \\ t = \frac{5}{10} = 0.5s$