Question

A stone is thrown vertically upward with an initial velocity of 40 m/s. Find the maximum height reached by the stone. What is the net displacement and the total 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. Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Answer 2

$\\ \purple{\bold {\large \underline{ \underline{Solution :- }}}}$

It is given that ,

• The initial velocity of the stone , u = 40m/s

We need to find the maximum height reached by the stone, the net displacement and the total distance covered by the stone.

At maximum height, Final velocity (v) = 0

Now, Using the equation of kinematics as :

$\\ \sf \: \: \: \: \: \underline{ \: \: {v}^{2} - {u}^{2} = 2ad \: \: }$

In this equation, d is the maximum height reached

$\\ \sf \: \: \: a = - g$

$\\ \sf \: \: \: \: \: d = \frac{ {u}^{2} }{2g }$

$\\ \sf \: \: \: \: \: d = \frac{ {40}^{2} }{2 \times 10 }$

$\\ \implies \: \sf \: \: \: \: \: d = \frac{ {1600}}{2 \times 10 }$

$\\ \implies \: \sf \: \: \: \: \: d = \frac{ {1600}}{20 }$

$\\ \implies \: \sf \: \: \: \: \: d = 80m$

As the stone goes up and came down. it means that the distance covered by the stone is 2d that is 160m. for this the displacement of the stone is the 0 as the initial position is equal to final position.