Question

• A stone is thrown in a vertically
upward direction with a velocity
of 5 ms. If the acceleration of the
stone during its motion is 10 ms
in the downward direction, what
will be the height attained by the
stone and how much time will it
take to reach there? ​

Answer 1

Explanation:

$\bigstar\:\:\rm\blue{GIVEN}\:\:\bigstar$

• $\bf\blue{Initial\:Velocity=5m/s^1}$

• $\bf\red{Final\:Velocity=0m/s}$

• $\bf\green{Acceleration=-10m/s^2}$

$\bigstar\:\:\rm\blue{To\:Find}\:\:\bigstar$

• The Height attend by stone and how much time will it take to reached there.

$\bigstar\:\:\rm\green{Formulae\:Used}\:\:\bigstar$

• ${\boxed{\tt{\blue{V=u+at}}}}$

Where,

V= Final Velocity

u= Initial Velocity

a= Acceleration

t= Time

• ${\boxed{\tt{\red{S=ut+\dfrac{1}{2}(a)(t)^2}}}}$

Where,

S= Distance

u= Initial Velocity

a= Acceleration

t= Time

Now,

$\implies\rm\blue{v=u+at}$

$\implies\rm\blue{0=5+(-10)×t}$

$\implies\rm\blue{-5=-10t}$

$\implies\rm\blue{t=\frac{\cancel{-5}}{\cancel{10}}}$

$\implies\rm\blue{t=0.5s}$

Thus, The time taken to reach there is 0.5s.

Now,

$\implies\rm\red{S=ut+\frac{1}{2}(a)(t)^2}$

$\implies\rm\red{S=5×0.5+\frac{1}{\cancel{2}}\times{\cancel{-10}\times{0.5}^2}}$

$\implies\rm\red{S=2.5+-1.25m}$

$\implies\rm\red{S=1.25m}$

Thus, The Distance travelled by Stone is 1.25m.