Question

HEYA ☺

A PROJECTILE IS FIXED AT AN ANGLE OF 30 DEGREE WITH THE HORIZONTAL WITH VELOCITY 10m/s . AT WHAT ANGLE WITH THE VERTICAL SHOULD IT BE FIRED TO GET MAXIMUM RANGE.☺
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Answer 1

Question.

A projectile is fixed at an angle of 30°with the horizontal with velocity 10 m/s . At what angle with the vertical should it be fired to get maximum range?

Solution.

Let the angle with the horizontal be θ.

• v = 10 m/s
• θ = 30°
• g = 10 m/s²

We know,

$\sf{Range=\dfrac{v^2sin2\theta}{g} }$

Putting the values :-

$\sf{\implies Range=\dfrac{(10)^2sin60^\circ}{10} }\\\\\sf{\implies Range=10\times \dfrac{\sqrt{3}}{2} }\\\\\sf{\implies Range=5\sqrt{3}}\\\\\sf{Taking\ \sqrt{3}\ as\ 1.7:}\\\\\sf{\implies Range=5\times1.7}\\\\\sf{\implies Range=8.5\ m}$

We get θ = 45° for maximum range.  

→ So angle with vertical = 90° - 45° = 45°

Again putting the values :-

$\sf{Range=\dfrac{v^2sin2\theta}{g} }\\\\\sf{\implies Max.\ range=\dfrac{(10)^2sin45^\circ}{10} }\\\\\sf{\implies Max.\ range=10\times \dfrac{1}{\sqrt{2}} }\\\\\sf{\implies Max.\ range=2\times5\times\dfrac{1}{\sqrt{2}} }\\\\\sf{\implies Max.\ range=5\sqrt{2}}\\\\\sf{Taking\ \sqrt{2}=1.4,}\\\\\sf{\implies Max.\ range=5\times1.4}\\\\\Large\boxed{\boxed{\sf{\implies Max.\ range=7\ m}}}$

Answer 2

Answer:

hloo...

.

Range = \frac{ v^{2}sin(2 \alpha ) }{g}

where α=angle with horizontal.

we get α=45° for maximum range.

So angle with vertical = 90° - 45° = 45°

v = 10m/s

Range = 10²×sin(60)/9.8 = 8.8m

maximum range = 10²×sin(90)/9.8 = 10.2m..

.

Have a nice day ...