Question

$\huge{\orange{\underline{\underline{Question:-}}}}$

When a particle is projected at some angle to the horizontal it has a range r and time of flight t1 if the same particle is projected with the same speed at some other angle to have the same range its time of flight is t2 then :-

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Answer 1

Answer:

3.) t1•t2 = 2R/g

Step by step explanations :

Method 1

Given that,

Maximum range of a projectile at two different angles at the same speed are equal.

We know that,

Maximum Range of a projectile at a vertical angle @

and Maximum range at angle (90 - @ ) are equal

given the time of flight at angle @ = t1

so,

t1 = (2usin@)/g

and time of flight at angle (90 - @) = t2

so,

t2 = 2usin(90 - @) /g

t2 = 2ucos@/g

Now,

t1 • t2

(2usin@/g)(2ucos@/g)

= (2u² 2sin@ cos@ )/g²

= 2u²sin2@/g² [2sin@ cos@ = sin2@]

= 2R/g

[R = u² sin2@/g]

so,

t1•t2 = 2R/g

_____________

Method 2

Refer to the attachment

Answer 2

$\huge\textbf{ t_1t_2 = \dfrac{2R}{g} }$

Option - C

$\huge{\textbf{\underline{As per Question}}}$

Let the projected velocity be u and range be R and time taken by the projected body be t1 and t2 with angle $\theta$.

• As we know in projectile body range is same if the angles is complementary.

Let the angle in first case be$\theta$ and second case be $(90-\theta)$

$\huge{\textbf{\underline{Solution}}}$

Time period of projectile is :-

$\mathsf{t_1 = \dfrac{2uSin\theta}{g}}$

$\mathsf{ t_2 = \dfrac{2uSin(90-\theta)}{g}}$

• multiply both equations.

$\mathsf{ t_1 \times t_2 = \dfrac{2uSin\theta}{g} \times \dfrac{2uSin(90-\theta)}{g}}$

$\mathsf{t_1 t_2 = \dfrac{2u Sin\theta\times 2u Cos\theta}{g^2}}$

$\mathsf{t_1 t_2 = \dfrac{4u^2 Sin\theta Cos\theta }{g^2}}$

• by arranging the values.

$\mathsf{t_1 t_2 = \dfrac{u^2 2 Sin\theta Cos\theta}{g} \times \dfrac{2 }{g}}$

$\mathsf{t_1 t_2 = R \dfrac{2}{g}}$

$\mathsf{t_1t_2 = \dfrac{2R}{g}}$