Question

For a given initial velocity the horizontal range of a body is 20m when the angle of projection is 60°. What is the horizontal range if it is thrown making an angle of 30° with horizontal?​

Answer 1

Explanation:

ACOORDING TO MATHEMATICAK RELATIONS WHEN A BALL IS THROWN INTO 60 DEGREE THEN DISTANCE IS 20 M

IF IT IS THROWN AT 30 DEGREE THEN IT SHOULD BE 10 M.

HOPE IT HELPS

Answer 2

Given :

• Acceleration due to gravity (g) = 10 m/s²
• Horizontal range (R) when angle of projection is 60° = 20 m
• Horizontal range (R) when angle of projection is 30° = ?

Answer:

First we need to find the initial velocity :

$\longrightarrow\:\:\tt R = \dfrac{2u^2\sin(\theta)\cos(\theta) }{g}$

$\longrightarrow\:\:\tt 20= \dfrac{2u^2\sin( {60}^{ \circ} )\cos({60}^{ \circ}) }{10}$

$\longrightarrow\:\:\tt 20 \times 10= 2 \times u^2 \times \dfrac{ \sqrt{3} }{2} \times \dfrac{1}{2}$

$\longrightarrow\:\:\tt \dfrac{200}{2} = u^2 \times \dfrac{ \sqrt{3} }{2} \times \dfrac{1}{2}$

$\longrightarrow\:\:\tt 100 = u^2 \times \dfrac{ \sqrt{3} }{4}$

$\longrightarrow\:\:\tt 4 \times 100 = u^2 \times \sqrt{3}$

$\longrightarrow\:\:\tt 400 = u^2 \times \sqrt{3}$

$\longrightarrow\:\:\tt u^2 = \dfrac{400}{ \sqrt{3} }$

Now, let's find the horizontal range for an inclined angle 30° :

$\longrightarrow\:\:\tt R =\dfrac{u^2\sin2(\theta) }{g}$

$\longrightarrow\:\:\tt R = \dfrac{ \frac{400}{ \sqrt{3} } \sin2( {30}^{ \circ} ) }{10}$

$\longrightarrow\:\:\tt R = \dfrac{ \frac{400}{ \sqrt{3} } \sin( {60}^{ \circ} ) }{10}$

$\longrightarrow\:\:\tt R = \dfrac{ \frac{400}{ \sqrt{3} } \times \frac{ \sqrt{3} }{2} }{10}$

$\longrightarrow\:\:\tt R = \dfrac{ \frac{400}{2} }{10}$

$\longrightarrow\:\:\tt R = \dfrac{ 200 }{10}$

$\green{\longrightarrow\:\:\tt R = 20 \: m}$