Question

(a) What is Archimedes principle?
(b) What is relative density?
(c) A ball is thrown vertically upwards with a velocity of $49\,m{s^{ - 1}}$. Find:
(i) The maximum height to which it rises.
(ii) The total time it takes to return to the surface of the earth. $\left( {g = 9.8\,m{s^{ - 2}}} \right)$.

Answer 1

Hint:Archimedes principle is the principle by using this principle the object is floating in the water. Ship floats in the water by using this Archimedes principle. Relative density is the ratio of the density of the substance to the density of a standard. By using the equation of motion, the height of the ball and the total time taken can be determined.

Useful formula:
The equation of motion is given by,
${v^2} - {u^2} = 2gh$
Where, $v$ is the final velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity and $h$ is the height of the object.
Acceleration due to change in velocity,
$v = u + at$
Where, $v$ is the final velocity, $u$ is the initial, $a$ is the acceleration and $t$ is the time taken.

Complete step by step solution:
(a) Archimedes principle:
The upward buoyant force that is exerted on a body which is immersed in a fluid, whether partially or fully submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the centre of mass of the displaced fluid. The value of thrust force is given by the Archimedes law which was discovered by Archimedes of Syracuse of Greece. When an object is partially or fully immersed in a liquid, the apparent loss of weight is equal to the weight of the liquid displaced by it.

(b) Relative density:
Relative density is also called specific gravity. The relative density can be calculated by dividing the density of the sample to the density of the standard liquid. For liquid, the standard density is the density of water. For gas, the standard density is the density of air.
Relative density of the fluid = Density of sample/Density of the standard fluid

(c) A ball is thrown vertically upwards with a velocity of $49\,m{s^{ - 1}}$.
Given that,
The initial velocity, $u = 49\,m{s^{ - 1}}$
The equation of motion is given by,
${v^2} - {u^2} = 2gh\,................\left( 1 \right)$
The final velocity of the ball is zero, because the ball reaches the maximum height the velocity is zero and then falls back. So, the final velocity $v = 0\,m{s^{ - 1}}$.
And then substituting the initial velocity and the acceleration due to gravity in the equation (1), then
${0^2} - {\left( {49} \right)^2} = 2 \times \left( { - 9.8} \right) \times h$
Here, the ball moves upwards, the acceleration due to gravity acts downwards, so the negative sign is included in the acceleration due to gravity.
By solving the above equation, then
$ - 2401 = - 19.6 \times h$
The negative sign on both sides, gets cancelled, and by rearranging the equation, then
$h = \dfrac{{2401}}{{19.6}}$
On dividing the above equation, then
$h = 122.5\,m$
Now,
Acceleration due to change in velocity,
$v = u + at\,................\left( 2 \right)$
Substituting the in initial velocity and the acceleration due to gravity in the equation (2), then
$0 = 49 - 9.8t$
Here, the ball moves upwards, the acceleration due to gravity acts downwards, so the negative sign is included in the acceleration due to gravity.
By solving the above equation, then
$ - 49 = - 9.8t$
By keeping, the time $t$ in one side and the other terms in other side, then
$t = \dfrac{{ - 49}}{{ - 9.8}}$
By cancelling the negative symbol and then dividing, then
$t = 5\,s$
The above equation shows the time taken by the ball to reach the maximum height.
Therefore, total time taken is $ \Rightarrow 2 \times 5\,s \Rightarrow 10\,s$
Thus, the total taken by the ball is $10\,s$.

Note:In question (c), the acceleration due to the gravitational force is taken as the negative, because the ball moves upwards, the acceleration due to the gravitational force acts towards the ground, so the negative sign is used.