Question

a man can throw a stone with initial speed of 10 m/s maximum range that can be achieved is

Answer 1

Answer:

R=u^2/g

=10×10/10

=10

Explanation:

hpe this wil help u a lot

Answer 2

Given:

A man can throw a stone with initial speed of 10 m/s.

To find:

Max range that can be achieved.

Diagram:

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Calculation:

General expression of range is :

$\boxed{ \bold{R = \dfrac{ {u}^{2} \sin(2 \theta) }{g} }}$

• "u" is initial velocity , $\theta$ is the angle of projection.

Now , for maximum range , value of sine must be max.

$\therefore \: \sin(2 \theta) = 1$

$\implies \: 2 \theta= {90}^{ \circ}$

$\implies \: \theta= {45}^{ \circ}$

So, max range obtained at 45° angle of Projection.

$\therefore \: R_{max} = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies \: R_{max} = \dfrac{ {u}^{2} \sin(2 \times {45}^{ \circ} ) }{g}$

$\implies \: R_{max} = \dfrac{ {u}^{2} \sin( {90}^{ \circ} ) }{g}$

$\implies \: R_{max} = \dfrac{ {u}^{2} \times 1 }{g}$

$\implies \: R_{max} = \dfrac{ {u}^{2} }{g}$

$\implies \: R_{max} = \dfrac{ {(10)}^{2} }{10}$

$\implies \: R_{max} = 10 \: m$

So, max range is 10 metres.