Question

A ball is projected with a velocity af 40 m/s
at the angle of 60° with the horizontal
cohat maximum will it rise ?
What will be its horizontal range?
(g= 10m/s²)​

Answer 1

Given :

initial velocity = 40m/s

angle of projection = 60°

To find :

The horizontal range.

Solution :

First we have to find the maximum height,

Maximum height (h) is given by,

$\boxed{ \sf h = \dfrac{ {u}^{2} {sin}^{2} \theta }{2g} }$

where,

• h denotes maximum height
• u denotes initial velocity
• g denotes gravity

substituting the given values,

$: \implies{ \sf h = \dfrac{ {(40)}^{2} {sin}^{2} 60 \degree }{2 \times 10} }$

$: \implies{ \sf h = \dfrac{ 1600 \times \dfrac{3}{4} }{20} }$

$: \implies{ \sf h = \dfrac{ 400 \times 3}{20}}$

$: \implies{ \sf h = 60 \: m }$

Horizontal range (R) is given by,

$: \implies \sf R = \dfrac{ {u}^{2} {sin}2 \theta }{g}$

$: \implies \sf R = \dfrac{ {(40)}^{2} {sin}2 (60 \degree) }{10}$

$: \implies \sf R = \dfrac{ 1600 \times \dfrac{ \sqrt{3} }{2} }{10}$

$: \implies \sf R = 160 \times \dfrac{ \sqrt{3} }{2}$

$: \implies \underline{\boxed{ \sf R = 80 \sqrt{3} \: m}}$

thus, the horizontal range is 80√3 m.

Answer 2

Given :-

Initial Velocity = 40m/s

Projection = 60°

To find :-

Horizontal Range

Correct explanation:-

$\longrightarrow{ \sf h = \dfrac{ {u}^{2} {sin}^{2} \theta }{2g} } \longleftarrow$

$\to{ \sf h = \dfrac{ {(40)}^{2} {sin}^{2} 60 \degree }{2 \times 10} }$

$\to{ \sf h = \dfrac{ 1600 \times \dfrac{3}{4} }{20} }$

$\to{ \sf h = \dfrac{ 400 \times 3}{20}}$

$\to{ \sf h = 60 \: m }$

Finding Horizontal range

$\to \sf R = \dfrac{ {u}^{2} {sin}2 \theta }{g}$

$\to \sf R = \dfrac{ {(40)}^{2} {sin}2 (60 \degree) }{10}$

$\to\sf R = \dfrac{ 1600 \times \dfrac{ \sqrt{3} }{2} }{10}$

$\to \sf R = 160 \times \dfrac{ \sqrt{3} }{2}$

$\sf \to{\green{ R = 80 \sqrt{3} \: m}}$

Horizontal range = 80√3 m.