Question

A stone is thrown vertically upwards with a velocity of 40 m/s.Find its position after 5 seconds

Answer 1

Answer:

75m above the projection point

Explanation:

use h=ut+(1/2)gt²

putting u=40m/s t=5s and g= -10m/s²

u easily obtain h=75m

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

u = 40 m/s.

t = 5 seconds.

g = - 10 m/s².

(As stone is travelling against the gravity)

$\huge{\bold{\underline{Explanation:-}}}$

From second kinematic equation,

$\large{\bold{S = ut + \frac{1}{2}at^2}}$

Substituting the values,

$\large{S = 40 \times 5 + \frac{1}{2} \times- 10 \times (5)^2}$

$\large{S = 200 - \frac{1}{2} \times 10 \times 25}$

$\large{S = 200 - \frac{1}{ \cancel{2}} \times \cancel{10} \times 25}$

Now,

$\large{S = 200 - 25 \times 5}$

$\large{S = 200 - 125}$

$\huge{\boxed{\boxed{S = 75 \: meters}}}$

So,the distance travelled after 5 seconds is 75m.

Additional formulas:-

$\large{ \star \: v = u + at}$

$\large{ \star \: v^2 - u^2 = 2as}$

$\large{ \star \: S_n = u + \frac{1}{2}a(2n - 1)}$