Question

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

DO SOLVE WITH THE DETAILED SOLUTION​

Answer 1

Explanation:

If an body thrown upward it has given initiall velocity

u=5m/s

a=-10m/s

v²-u²=2as

v²-u²=2gh

v² is directly proportional to h

v²-5m/s²=2×-20m/s×h

v²-25m/s²=-40m/s²×h

v²=-40m/s²+25m/s²×h

v²=-15m/s²×h

v=√-15m/s²

h=√-15m

Answer 2

$\maltese$ Answer $\maltese$

$\star$ Given:

⇒ Initial Velocity ( u ) = 5 m/s

⇒ Acceleration ( a ) = 10 m/s²

⇒ Final Velocity ( v ) = 0 m/s

$\star$ To Find:

⇒ Height ( S )

⇒ Time ( t )

$\star$ Solution:

$\checkmark$ Equation of Motion

$\orange{\boxed{\boxed{\rm v^{2}=u^{2}+2aS}}}$

Given that,

→ Initial Velocity ( u ) = 5 m/s

→ Final Velocity ( v ) = 0 m/s

Acceleration ( a ) = 10 m/s²

Since, the Acceleration is in downward direction

Thus, The stone has a Negative Acceleration

→ Acceleration ( a ) = -10 m/s²

We are asked to Find Height ( S )

Therefore,

From the Given Equation of Motion

We get,

$\blue{\implies \rm S=\dfrac{v^2-u^2}{2(a)}}$

Substituting the above values in the Formula

We get,

$\implies \rm S=\dfrac{0^2-5^2}{2(-10)} \ m$

$\implies \rm S=\dfrac{-25}{-20} \ m$

$\rm \implies S=1.25 \ m$

Hence,

Height ( S ) = 1.25 m

We are also asked to Find the Time ( t ) to reach

Using another Formula

$\checkmark$ Equation of Motion

$\red{\boxed{\boxed{\rm v=u+at}}}$

Given that,

→ Initial Velocity ( u ) = 5 m/s

→ Final Velocity ( v ) = 0 m/s

→ Acceleration ( a ) = -10 m/s²

Therefore,

Substituting the above values in the Formula

We get,

$\implies \rm 0=5+(-10\;t)$

$\implies \rm -10\;t=-5$

$\implies \rm t=\dfrac{5}{10} \ sec$

$\rm \implies t=0.5 \ sec$

Hence,

Time ( t ) = 0.5 seconds

So Finally,

Height ( S ) = 1.25 m

Time ( t ) = 0.5 seconds