Question

A projectile has the same range r for two angle of projection. If t1 and t2 be the time of flight in the two cases, then velocity of projection of projectile is

Answer 1

projectile can have same range for angles of projection ϴ and 90-ϴ

=> T1=2Usinϴ/g T2=2Usin(90-ϴ)/g=2Ucosϴ/g

T1*T2=4U^2sinϴcosϴ/g^2=2R/g [R is range]

so it is directly proportional to range of projectile

Answer 2

Answer:

Explanation:

Given

Range of projectile is R

let \theta be the launch angle thus its complimentary angle is 90-\theta

$t_1 is time for \theta$

$t_2 is time for 90-\theta$

let u be the initial velocity

$R=\frac{u^2sin2\theta }{g}$

time of flight of Projectile

$t_1=\frac{2usin\theta }{g}$

$u\sin \theta =\farc{gt_1}{2}$---1

$t_2=\frac{2usin(90-\theta )}{g}$

$u\cos \theta =\farc{gt_1}{2}$---2

squaring and adding 1 & 2 we get

$u^2=\left ( \frac{gt_1}{2}\right )^2+\left ( \frac{gt_2}{2}\right )^2$

$u=\sqrt{\left ( \frac{gt_1}{2}\right )^2+\left ( \frac{gt_2}{2}\right )^2}$