Question

A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10 m/s^2 find the maximum height reach by the stone and what is the net displacement and distance covered by the stone.​

Answer 1

The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Step-by-step explanation:

Final velocity v = 0

Initial velocity u = 40m/s

We know that,

Using equation of motion

`v^(2) =u^(2) +2gh`

`0-40^(2) =2` ×
`10` ×
`h`

The maximum height is:

`h=80`
`m`

The stone will reach at the top and will come down

Therefore, the total distance will be:

`s=h_(1) +h_(2)`

`s=80m-80m=160m`

The net displacement is:

`D=h_(1) -h_(2)`

`D=80m-80m=0`

Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

hope this helps.....

Answer 2

Step-by-step explanation:

u=40

v=?

h=?

v²-u²=2gs

0²-40²=2×10×s

160=20s

s=160/20

=80m/s

total distance= upward distance ×downward distance

=80+80

=160m

total displacement=0 because u and v is the same.