Question

(b) A projectile with initial speed u, can have the same range R for two angles of
projection. If t1 and t2 be the time of flight in the two cases, then find the value of
T1T2 in terms of R

Answer 1

Answer:

Given:

A Projectile with speed u , has same range R , for 2 different time period.

To find:

t1 × t2 = ? (in terms of R)

Calculation:

For constant Projection velocity and same range , this condition is only possible only when a Projectile has been thrown at complementary angles.

Let one angle be θ and the

other be (90 - θ).

Calculation:

Time for 1st Projection be t1

$t1 = \dfrac{2u \sin( \theta) }{g}$

Time for 2nd Projection be t2

$t2 = \dfrac{2u \sin(90 \degree - \theta) }{g}$

$= > t2 = \dfrac{2u \cos( \theta) }{g}$

Now as per question :

$\therefore \: t1 \times t2$

$= > \dfrac{2u \sin( \theta) }{g} \times \dfrac{2u \cos( \theta) }{g} \:$

$= > ( \dfrac{2}{g}) \times \dfrac{ {u}^{2} \times 2 \sin( \theta) \cos( \theta) }{g}$

$= > ( \dfrac{2}{g} ) \times \dfrac{ {u}^{2} \times \sin(2 \theta) }{g}$

$= > \dfrac{2R}{g}$

So final answer is:

$\boxed{ \large{ \red{t1 \times t2 = \frac{2R}{g}}}}$

Answer 2

☆☆Refer to the attachment please