Question

one stone is thrown downward with an inital velocity of 1 m/sec while the second stone is drop simultaneously. what will be the separation between them after 18 sec.(g=10 m/sec^2)

Answer 1

STONE 1 :-

u is 1 m/sec

a is equal to acceleration due to gravity that is 10 m/sec²

t is 18 seconds

so,

we can find distant travelled

$s = ut + \frac{1}{2}a {t}^{2} \\ \\ \\s = 1 \times 18 + 5 \times {(18)}^{2} \\ \\ \\s = 18 + 1620 \\ \\ \\ s = 1638m$

STONE 2:-

u is zero

a is 10 m/sec²

time is 18 seconds

finding the distance travelled:-

$s = ut + \frac{1}{2}a {t}^{2} \\ \\ \\s = 5 \times {(18)}^{2} \\ \\s = 5 \times 324 = 1620m$

Hence,

The separation between them is equal to

$= 1638 - 1620 \\ \\ \\ = 18 \: metres$

In such cases we can do it directly

the direct formula is :-

$= ut$

It is for the stone thrown with some initial velocity u and time of checking the separation is t

Answer 2

$\huge{\underline{\orange{\mathfrak{Answer :-}}}}$

Given :-

Initial velocity(u) = 1m/s

Time taken(t) = 18 sec

Acceleration due to gravity(g) = 10m/s²

Soution :-

Case 1 :-

We know that,

$\huge{\red{\boxed{\boxed{\bf{s = ut + \frac{1}{2}at^{2}}}}}}$

_________________[Put Values]

${ \sf{s \: = 1(18) \: + \frac{1}{2} \times \:10 \times {18}^{2} }}$

s = (18 + 10 * 324) / 2

s = (18 + 1620)

s = 1638 m

$\huge{\red{\boxed{\boxed{\bf{s = 1638m}}}}}$

$\rule {200}{2}$

Case 2 :-

Given :-

Initial velocity(u) = 0 m/s

Acceleration due to gravity(g) = 10m/s²

Time taken(t) = 18 sec

We know that,

$\huge{\red{\boxed{\boxed{\bf{s = ut + \frac{1}{2}at^{2}}}}}}$

_________________[Put Values]

s = 0(18) + 5 * (18)²

s = 5 * (18)²

s = 5 * 324

s = 1620 m

$\huge{\red{\boxed{\boxed{\bf{s = 1638m}}}}}$

$\rule {200}{2}$

Seperation between them = 1638 - 1620 = 18 m

$\huge{\red{\boxed{\boxed{\bf{separation\: between \: them = 18 m}}}}}$