Question

Question=the stone is thrown vertically upward with an initial velocity of 40m/s taking g=10m/s² find the maximum height reached by the stone. what is the net displacement and the total distance covered by the stone ?​

Answer 1

Answer:

Hmax=80m

net displacement=0 bcz it finally end up from where it started

net distance=160m

Answer 2

GiveN :-

• A stone is thrown vertically upward with an initial velocity of 40m/s.
• Take the value of g as 10 m/s² .

To FinD :-

• The maximum height reached by the stone.
• The net displacement and total distance covered by the stone.

SolutioN :-

The initial velocity of the stone is 40m/s². And we need to take the value of g as 10m/s² . We need to find firstly maximum height . We know maximum height as ,

$\red{\bigstar}\underline{\textsf{ Using the formula of Maximum height:- }}$

$\sf:\implies \pink{ Height_{(maximum)}=\dfrac{u^2}{2g }}\\\\\sf:\implies H_{(max)}= \dfrac{(40 m/s)^2}{2(10 m/s^2)} \\\\\sf:\implies H_{(max)}= \dfrac{ 1600 m^2/s^2}{20 m/s^2}\\\\\sf:\implies \underset{\blue{\sf Required\ Answer}}{\underbrace{\boxed{\pink{\frak{ Height _{(maximum)}= 80 m }}}}}$

This maximum height is only the Displacement and the Distance travelled by the stone is 80m+80m = 160m .