Question

Falling objects question?

Answer 1

Applicable kinematic expressions are

`h=h_0+ut+1/2g t^2` ......(1)
where `h` is height at time `t`, `h_0` is height at `t=0`, `u` is initial velocity and `g=9.80\ ms^-2` is acceleration due to gravity.
`v=u+g t` .......(2)
where `v` is velocity at time `t`

1. Let us take origin at the ground level.
2. As the stone is thrown upwards, along the `+ve`-`y` axis and gravity is acting in the downwards direction, it is taken as `-ve`. Our equations become

   `h=50.0+20.0t-4.90 t^2` .........(3)
   `v=20.0-9.80 t` ..............(4)

(A) When stone reaches maximum height its velocity becomes `0`. Using (4) we get

`0=20.0-9.80 t`
`=> t=20.0/9.80=2.04\ s`, rounded to two decimal places.

(B) Insert calculated value of `t` in part (A) above in (3)

`h_max=50.0+20.0xx2.04-4.90 xx(2.04)^2`
`h_max=50.0+40.8-20.4=70.4\ m`

(C) Ignoring friction, velocity of stone when it reaches same height from it was thrown will be equal and opposite to the initial velocity of projection.

`v=-20.0\ms^-1`

(D) Velocity of stone at `t=5.00\ s` can be found using (4)

`v=20.0-9.80 xx5.00=-29.0\ ms^-1`

Position of stone at `t=5.00\ s` is found using (3)

`h=50.0+20.0xx5.00-4.90 xx(5.00)^2=27.5\ m`

(E) When the stone strikes ground its height `h=0`. Using (3) to find total time of flight `T`

`0=50.0+20.0T-4.90 T^2`
`=>4.90 T^2-20.0T-50.0`

Solving the quadratic we get

`T=(20+-sqrt(400-4xx4.90xx(-50.0)))/(2xx4.90)`
`=>T=(20+-sqrt(400+980))/(9.8)`
`=>T=(20+-37.15)/9.80`
`=>T=5.83\ s`,

We have ignored the `-ve` root as time can not be `-ve.`

Velocity of stone when it just strikes ground is found by inserting value of `t=T` in (4)

`v=20.0-9.80 xx5.83=-37.1\ms^-1`