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

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.

Answer 2

Answer:

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

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.....