Question

3. A stone is thrown in a vertically upward direction with a vilocity of 5 m per sec if the acceleration of stone during its motion is 10m per sec square in the downward what will be the hight attend by the stone and how much time will it take to reach their​

Answer 1

Given :-

• A stone is thrown in a vertically upward direction with a velocity of 5 m/s .

• The acceleration of stone during its motion is 10 m/ s² in the downward.

To Find :-

• The height attend by stone and how much time will lt take to reach there.

Solution:-

• To calculate height and time at first we have to set up equation as per given clue. We will find out time and height with the help of law of motion.

Formula Used :-

==> v² = u² + 2 a s

==> v = u + at

=>Let's find out time :-

=> Final velocity (v) = 0

=> Intial velocity ( u) = 5 m/s

=> acceleration (a) = -10 m/s²

=> v= u + at

==> 0 = 5 + (-10)×t

==> 0 = 5 - 10t

==> 10t = 5 => t = 0.5 s

=> Let's height attend by stone :-

=> Final velocity (v) = 0

=> Initial velocity (u) = 5 m/s

=> time taken (t) = 0.5 s

=> acceleration (a) = -10 m/s²

==> v² = u² + 2 a s

• Here s is height

==> v² = u² + 2 a h

==> 0² = 5² + 2 × (-10) × h

==> 0 = 25 - 20 h

==> 20 h = 25

==> h = 25/20

==> h = 1.25 m

Answer 2

Answer:

Given :-

• A stone is thrown in a vertically upward direction with a velocity of 5 m/s. The acceleration of stone during its motion is 10 m/s² in the downward.

To Find :-

• What is the height attend by the stone.
• How much time will it take to reach.

Formula Used :-

$\clubsuit$ First equation of motion :

$\longmapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Third equation of motion :

$\longmapsto \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2as}}}$

where,

• s = Distance Travelled
• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time

Solution :-

First, we have to find the height attend by the stone :

$\mapsto$ The height attend by the stone = s m

Given :

• Final Velocity (v) = 5 m/s
• Initial Velocity (u) = 0 m/s
• Acceleration (a) = 10 m/s²

According to the question by using the formula we get,

$\dashrightarrow \sf (5)^2 - (0)^2 =\: 2(10)s$

$\dashrightarrow \sf 25 - 0 =\: 2 \times 10 \times s$

$\dashrightarrow \sf 25 - 0 =\: 20s$

$\dashrightarrow \sf 25 =\: 20s$

$\dashrightarrow \sf \dfrac{\cancel{25}}{\cancel{20}} =\: s$

$\dashrightarrow \sf \dfrac{\cancel{5}}{\cancel{4}} =\: s$

$\dashrightarrow \sf 1.25 =\: s$

$\dashrightarrow \sf\bold{\red{s =\: 1.25\: m}}$

$\therefore$ The height attend by the stone is 1.25 m .

Now, we have to find the time will it take to reach :

Given :

• Final Velocity (v) = 5 m/s
• Initial Velocity (u) = 0 m/s
• Acceleration (a) = 10 m/s²

According to the question by using the formula we get,

$\implies \sf 5 =\: 0 + 10(t)$

$\implies \sf 5 - 0 =\: 10t$

$\implies \sf 5 =\: 10t$

$\implies \sf \dfrac{5}{10} =\: t$

$\implies \sf 0.5 =\: t$

$\implies \sf\bold{\red{t =\: 0.5\: seconds}}$

$\therefore$ The time will it take to reach is 0.5 seconds.