Question

an objects through vertically upwords reches a height of320m . what was its intial velocity? how long will the object taken to come back Earth?assumeg=10m/s2​

Answer 1

Answer:

• Initial velocity, u = 80 m/s
• Time taken = 8 seconds

An object thrown vertically upward reaches a height of h = 320 m , while acceleration due to gravity is g = 10 m/s²

Now, we have to find the following quantities

• Initial velocity, u
• Time taken to come back

We have,

• Final velocity, v = 0 m/s as the object will stop at the topmost position which is at a height of 320 m.

Using the third equation of motion, we have

⇒ 2as = v² - u²

⇒ 2 × -10 × 320 = 0 - u²

[ Displacement or distance travelled by the object is given to be 320 m while you may observe that g here, is taken as negative that's because g is along the opposite direction of the motion of object. ]

⇒ -6400 = -u²

⇒ u² = 6400

⇒ u = 80 m/s

Now, we have to find the time taken by the object to come back, which is equal to the time taken by it to reach there,

We have,

• v = 80 m/s
• u = 0 m/s
• g = 10m/s

Now, Final velocity is equal to the initial velocity at which the object was thrown because this is free fall. While at the topmost position, the velocity is 0 and g is taken as positive as it is in direction of motion.

Using the first equation of motion,

⇒ v = u + at

⇒ 80 = 0 + 10t

⇒ t = 80/10

⇒ t = 8 seconds.

Answer 2

Answer:

Given :-

• An object through vertically upwards reaches a height of 320 m.

To Find :-

• What is the initial velocity.
• How long will the object taken to come back to the Earth.

Solution :-

First, we have to find the initial velocity :

As we know that :

$\clubsuit$ Newton third equation of motion :

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

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance travelled

Given :

• Final velocity = 0 m/s
• Distance travelled = 320 m
• Acceleration = - 10 m/s²

According to the question by using the formula we get,

$\implies \sf {0}^{2} =\: {u}^{2} + 2 \times (- 10) \times 320$

$\implies \sf 0 =\: {u}^{2} - 20 \times 320$

$\implies \sf 0 =\: {u}^{2} - 6400$

$\implies \sf \cancel{-} {u}^{2} =\: \cancel{-} 6400$

$\implies \sf {u}^{2} =\: 6400$

$\implies \sf u =\: \sqrt{6400}$

$\implies \sf\bold{\red{u =\: 80\: m/s}}$

$\therefore$ The initial velocity is 80 m/s .

$\rule{150}{2}$

Now, we have to find the how long will the object taken to come back to the Earth :

As we know that:

$\clubsuit$ Newton first law of motion :

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

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time

Given :

• Final Velocity = 0 m/s
• Initial Velocity = 80 m/s
• Acceleration = - 10 m/s²

According to the question by using the formula we get,

$\implies \sf 0 =\: 80 + (- 10) \times t$

$\implies \sf 0 =\: 80 - 10t$

$\implies \sf \cancel{-} 80 =\: \cancel{-} 10t$

$\implies \sf 80 =\: 10t$

$\implies \sf \dfrac{8\cancel{0}}{1\cancel{0}} =\: t$

$\implies \sf \dfrac{8}{1} =\: t$

$\implies \sf 8 =\: t$

$\implies \sf\bold{\red{t =\: 8\: seconds}}$

$\therefore$ The time taken to come back to the Earth is 8 seconds .