Question

A stone is thrown from upward it's initial velocity is 36 m/ s . Find how distance covered ?

Answer 1

Ball is Thrown UPWARD

Intial Velocity (u)=36 m/s

Final Velocity on Maximum Height=0m/s

Final Velocity (v)=0 m/s

Accerlation due to Gravity(a)=-10 m/s

(- Accerlation because it is Working Against Gravity)

Distance (s)=?

Use Equation of Motion

$\underline{\mathbf{ {v}^{2} = {u}^{2} + 2 \times a \times s}}$

Putting Value

$\mathbf{ {0}^{2} = {36}^{2} + 2 \times - 10 \times s} \\ \\ \mathbf{0 = 1296 + (- 20s)} \\ \\ \mathbf{-20s = - 1296} \\ \\ \mathbf{s = \frac{ \cancel{ - 1296}}{ \cancel{ - 20}} } \\ \\ \mathbf{ s = 64.8}$

$\boxed{\mathbf{\huge{Distance=64.8\:m}}}$

Answer 2

u = 40 m/s

As the stone is thrown upward the acceleration due to gravity is to be taken negative.

g = - 10 m/s2

v2 - u2 = 2as

For free fall we can write this equation as,

v2 - u2 =2gh

As the stone reaches the maximum height its final velocity v =0

Thus,

0 - (40)2 = 2× (-10) × h

- 1600 = -20 × h

h = 80 m

So, the maximum height to which the stone can reach is 80 m.

The total distance covered by the stone = 80 + 80 = 160 m

And as th stone comes back to its initial position the displacement of the stone = 0

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