Physics : asked on fanny1211
 22.11.2021

The acceleration of free fall on a certain planet is 8.0 m s-2. An object gets dropped from a height and hits the ground after 1.5 s. From what height must the object have been dropped?

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09.07.2023, solved by verified expert
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Answer:Height from which the object is dropped is 9m

We are given that:

The acceleration of free fall on a certain planet is 8m/s². An object is dropped from the height and hits the ground after 1.5s

We need to find:

Height from which the object was dropped.

Using equations of motion:

➙ v² - u² = 2as

➙ v = u + at

here,

u = initial velocityv = final velocitya = accelerations = displacementt = time taken Solution:

From the given information, we can conclude that:

initial velocity of the object is 0 m/sacceleration due to gravity = 8 m/s²time taken = 1.5 seconds

Now, using equation of motion:

➝ v = u + at

➝ v = 0 + ( 8 × 1.5 )

➝ v = 0 + 12 .0

➝ v = 12 m/s

final velocity of object = 12 m/s

For, acceleration :

➝ v² - u² = 2as

➝ ( 12 )² - ( 0 )² = 2 × 8 × s

➝ 144 - 0 = 16s

➝ 16 s = 144

➝ s = 144 / 16

➝ s = 9m

Displacement = 9 m

The acceleration of free fall on a certain planet, №18010764, 22.11.2021 10:40 Since, we take displacement in terms of distance in some places, therefore we can say that the height from which the object was dropped is 9 meters .

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Physics
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P Answered by Master
Answer:Height from which the object is dropped is 9m

We are given that:

The acceleration of free fall on a certain planet is 8m/s². An object is dropped from the height and hits the ground after 1.5s

We need to find:

Height from which the object was dropped.

Using equations of motion:

➙ v² - u² = 2as

➙ v = u + at

here,

u = initial velocityv = final velocitya = accelerations = displacementt = time taken Solution:

From the given information, we can conclude that:

initial velocity of the object is 0 m/sacceleration due to gravity = 8 m/s²time taken = 1.5 seconds

Now, using equation of motion:

➝ v = u + at

➝ v = 0 + ( 8 × 1.5 )

➝ v = 0 + 12 .0

➝ v = 12 m/s

final velocity of object = 12 m/s

For, acceleration :

➝ v² - u² = 2as

➝ ( 12 )² - ( 0 )² = 2 × 8 × s

➝ 144 - 0 = 16s

➝ 16 s = 144

➝ s = 144 / 16

➝ s = 9m

Displacement = 9 m

\dag Since, we take displacement in terms of distance in some places, therefore we can say that the height from which the object was dropped is 9 meters .

Physics
Step-by-step answer
P Answered by Master

Answer:

see below.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

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