Understand the Acceleration Due to Gravity Formula and More!

How acceleration is equal to $d^{2}x/d{t}^2$

What is the formula for calculating wave velocity?

Find the acceleration due to gravity on the surface of mercury.

What is meant by the statement that acceleration due to gravity is 9.8m/s2?

What is rate of change of acceleration called?

Hello, Plainmath
Help me to solve this
The height of a helicopter above the ground is given by $h=3.25t^3$, where h is in meters and t is in seconds. At t = 2.25 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

The acceleration due to gravity at the surface of the moon is $1.67m/se{c}^2$. If the radius of the moon is $1.74\times {10}^{6}m$. Find the mass of the moon. ( Given: $G=6.67\times {10}^{-11}N{m}^2/k{g}^2$)

Proof of a formula relating launch velocity of projectile to times of ascent and descent
I came across this formula in my textbook stating that when a body is projected vertically upwards and the time corresponding to height $H$ while ascending and descending are $t_1$ and $t_2$ respectively, when the velocity of projection is $\frac{g(t_1+t_2)}{2}$ where $g$ is acceleration due to gravity.
Can anyone please help me in deriving this formula? I know that in case of vertical motion upwards the initial velocity will be gt1. But I don't know how to proceed further. Please help.

Why does gravity decrease as we go down into the Earth?
We all know that gravity decreases as the distance between the two increases. Hence
$F=G\frac{Mm}{r^2}.$
Hence the acceleration due to gravity
$g=\frac{F}{m}=G\frac{M}{r^2}$
increases as $r$ decreases. Then why does it decrease as we go deep into the earth?

How does the gravitational acceleration on a surface of planet change, if its radius shrinks while retaining the same mass?
If the radius of the earth were to shrink by $1$% while its mass remained the same,
then how would the acceleration due to gravity on the earth's surface increase by $2$%?

What will happen to a pendulum clock carried an astronaut on orbit around the earth?
Which one of the two following descriptions is the correct one:
1. a pendulum clock carried by an astronaut orbiting the earth will not swing because there's no gravity and the pendulum is in free fall.
2. a pendulum clock will swing at a reduced rate: the periodic time will increase because the acceleration due to gravity "g" is reduced.

Which part of a massless wire will break at the breaking stress?
Consider an ideal massless wire of length $L$, uniform cross-sectional area $A$, Young's modulus $Y$ suspended from the ceiling, with a load of weight $W$ suspended at the end. There is no variation in the value of acceleration due to gravity $g$ along the length of the wire. Any other factors that vary in real life are also constant here.
When the weight is just enough to produce the breaking stress in the wire, the wire breaks (fractures) at a point.
Which point will the wire break at? Is it entirely random (anywhere along the length) or is there a definite point (like the midpoint)?

How to calculate the increase in temperature due to drop?
Question- Calculate the rise in temperature in celcius in a bucket of water after it is dropped from 50 m where acceleration due to gravity is 10.
I know that I need to find the amount of energy absorbed and then find out the increase in temperature using the specific heat of water but cant do it as the mass isn't mentioned. This is a question in the book of my coaching center. I am a seventh grader. Thanks.

Rotation of free to rotate rigid bodies
I recently came across this statement in a book that the line joining the point of suspension and center of mass of a body should be parallel to the acceleration due to gravity(g or more precisely, g effective). Why is this so? Can anyone please explain this to me?

Free falling body's deceleration
Ok, a body is falling with a velocity $70\sqrt{2}m{s}^{-<!--\; -\; -->1}$ and its final velocity must be 0 $m{s}^{-<!--\; -\; -->1}$ after falling a further 200 $m$. What is the acceleration required to bring it to a stop after $200m$?
Using the formula
$v^2=v_0^2+2ax$
and substituting in the known quantities gives
$0=9800+(2a\times <!--\; \times \; -->200).$
Rearranging for $a$ gives
$\begin{array}{rl}a& =-<!--\; -\; -->\frac{9800}{400}\\ \\ & =-<!--\; -\; -->24.5m{s}^{-<!--\; -\; -->2}.\end{array}$
Ok that's easy enough and that's the given answer, but should we not account for $g$ in the answer also?
I mean the acceleration due to gravity hasn't vanished. 'It wants' to speed the body up, so shouldn't the deceleration be $(24.5+g)m{s}^{-<!--\; -\; -->2}$?

Definition of weight
As per CGPM the official definition of weight is
"The weight of a body is the product of it's mass and acceleration due to gravity."
When I searched for gravity it defined gravity as the fundamental force of attraction acting between any two bodies with non-zero mass. So should not the definition of weight contain "maximum value of acceleration due to gravity?"

How is weight $W=0$?
One of my physics teachers told me that in the formula of calculating weight i.e. $W=mg$ is not equal to 0 until and unless you're travelling close to the speed of light. She added, that the acceleration due to gravity, $g$ is also never equal to 0 because for that either mass of Earth has to be 0 or the square of the radius infinite. My question is, if $m$ and $g$ are both non-zero, then how is weight equal to 0?

How can there be acceleration but no velocity at instant of maximum height when object projected upwards?
When we project an object vertically upwards its velocity at maximum height consider to be zero but still, it has acceleration due to gravity i.e $9.8\mathrm{m}/{\mathrm{s}}^2$. But mathematically we know that acceleration is the first derivative of velocity with respect to time $(a=dv/dt)$. So mathematically if velocity is zero there can not be any acceleration but this case contradict this equation. We can not even say that its direction is changing as the direction has no meaning without magnitude. So how this situation can be explained mathematically?

What forces does a person experience after falling from a sixth storey building?
If someone falls from a sixth storey building, with what force will he land? By this I mean to say that if he weighs 50 kg and was initially at rest before falling, which implies it was a free fall. We can easily calculate the velocity and momentum as we know the acceleration due to gravity... but with what force he will land?

Why does the normal force go down in an downward accelerating elevator?
When standing in a stationary elevator the force due to gravity and the normal force cancel each other out hence the net force is 0. When the elevator starts accelerating downwards there's a net force in the downward direction meaning that the body in the elevator now experiences three forces, the force due to gravity,the force that is causing the acceleration of the whole elevator and the normal force, accordingly the downward force is now of a greater magnitude than that of the normal force the body experiences. Then my question is, why does the normal force need to reduce if the downward force is still greater than the normal force even if it doesn't decrease. The body feels the force of gravity and the force which causes the elevator to accelerate these two are greater than the normal force and therefore the elevator accelerates in the downwards direction, why should the normal force get smaller?