Question

Two particles are projected with same velocity but at angles of projection 35° and 55º. Then their horizontal ranges are in the ratio of
(A) 1:2
(B) 2:1
(C) 1:1
(D) 4:1​

Answer 1

Given :-

• Two particles are projected with the same velocity , Let u

• Angles of projection of two particles are 35° and  55° ,Let us name them α and β respectively.

To find :-

• Ratio of their horizontal ranges

Knowledge required :-

• Formula to calculate horizontal range

$\orange{\bigstar}\boxed{\sf{R=\dfrac{u^2\;sin\;2\theta}{g}}}$

(where R is the horizontal range , u is the initial velocity of particles , g is acceleration due to gravity and θ is the angle of projection)

Solution :-

Let, ranges of two particles be

R₁ and R₂

then,

Calculating Range of particle 1

Using formula to calculate range

$\implies\sf{R_1=\dfrac{u^2\;sin\;2\alpha}{g}}\\\\\\\implies\sf{R_1=\dfrac{u^2\;sin\;2(35^{\circ})}{g}}\\\\\\\implies\red{\sf{R_1=\dfrac{u^2\;sin\;70^{\circ}}{g}}}$

Calculating Range of particle 2

Using formula to calculate Range

$\implies\sf{R_2=\dfrac{u^2\;sin\;2\beta}{g}}\\\\\\\implies\sf{R_2=\dfrac{u^2\;sin\;2(55^{\circ})}{g}}\\\\\\\implies\red{\sf{R_2=\dfrac{u^2\;sin\;110^{\circ}}{g}}}$

Calculating the Ratio between their Ranges

$\implies\sf{\dfrac{R_1}{R_2}}\\\\\\\implies\sf{\dfrac{R_1}{R_2}=\dfrac{u^2\;sin\;70^{\circ}}{g}\;\div \;\dfrac{u^2\;sin\;110^{\circ}}{g}}\\\\\\\implies\sf{\dfrac{R_1}{R_2}=\dfrac{u^2\;sin\; 70^{\circ}}{g}\times \dfrac{g}{u^2\;sin\;110^{\circ}}}\\\\\\\implies\sf{\dfrac{R_1}{R_2}=\dfrac{sin\;70^{\circ}}{sin\;110^{\circ}}}\\\\\\\sf{since,\;sin\;70^{\circ}=sin\;110^{\circ}\;\;; therefore,}\\\\\\\implies\boxed{\boxed{\large{\red{\sf{R_1\;:\;R_2=1\;:\;1}}}}}$

Hence,

Correct option is (C) 1 : 1

The Ratio of horizontal ranges of two particles is 1 : 1 .

$\underline{\rule{205}3}}$

Let us prove how

 sin 70° = sin 110°

so,

Taking LHS

LHS = sin 70°

LHS = sin ( 180° - 110°)

using : sin ( 180 - A ) = sin A

LHS = sin 110° = RHS

Proved.

$\underline{\rule{205}3}}$