Question

A rock is dropped from a height of 2.5 meters. What is its velocity when it reaches the ground?

Answer 1

The velocity of the rock when it reaches the ground is `7 m/s`.

Explanation:

The formula we are using is `V_"2"^2=V_"1"^2+2aDeltad`
=> Where `V` is the velocity in `m/s`.
=> Where `a` is acceleration in `m/s^2`.
=> Where `Deltad` is the change in displacement in `m`.

So now we just plug in the numbers.

`V_"2"^2=V_"1"^2+2aDeltad`

`=0+2(9.8)(2.5)`

`=sqrt49`

`=7`

The velocity of the rock when it reaches the ground is `7 m/s`.

Hope this helps :)

Answer 2

The velocity of the rock when it hits the ground is `7"m"/"s"`.

Explanation:

The kinematic equation required to answer this question is:

`v_f^2=v_i^2+2ad`, where `v_f` is the final velocity, `v_i` is the initial velocity, `a` is acceleration, which is that of gravity, so we use `g` instead, and `d` is displacement.

First determine what information is given or known, and what is unknown.

Given/Known
`v_i="0 m/s"`
`g=-9.8 "m/s"^2"`
`d="-2.5 m"`

Unknown
`v_f` (the velocity when it reaches the ground)

Equation

`v_f^2=v_i^2+2gd`

Substitute the known/given values into the equation.

`v_f^2=0+2*-9.8"m"/"s"^2*-2.5"m"`

`v_f^2=49"m"^2/"s"^2`

`v_f=sqrt(49"m"^2/"s"^2)`

`v_f=7"m"/"s"`