Two particles A and B are projected from point o with equal speeds they both hit the point P

Ratio of their times of flights is 4:5

Ratio of their maximum hights is 9:25

Ratio of their minimum speeds during flight is 4:3

The bisector of the angle between their directions of projection makes 37º with horizontal.

Text Solution

`3(1-cos 2alpha)=8sin 2beta`Range of A=maximum height of B Maximum vlaue of `beta` is `(1)/(2)sin^(-1)(3)/(4)`Maximum horizontal range of A `=(u^(2))/(g)` and this occurs when `beta=(1)/(2)sin^(-1)(3)/(8)`

Answer : a,c,d

Solution : `(u^(2)sin^(2)alpha)/(2g)=(2)/(3)(u^(2)sin 2beta)/(g)` <br> `sin^(2)alpha=(4)/(3)xx2sin beta.cosbeta` <br> `((1-cos2alpha))/(2)=(4)/(3) sin 2beta` <br> `beta_("max")=sin^(-1)((3)/(4))`

Text Solution

Solution : Time of flights of `A and B` <br> `T_A = (2 u sin (prop + beta))/(g cos beta)` and <br> `T_B = ( 2 u sin (prop - beta))/(g cos beta)` <br> `(T_A)/(T_B) = (sin(prop + beta))/(sin(prop - beta)) = cot theta` (given) <br> `rArr (cos theta)/(sin theta) = (sin (90 - theta))/(sin theta) = (sin (prop + beta))/(sin(prop - beta))` <br> We can write, `prop + beta = 90 - theta` ...(i) <br> and `prop - beta = theta` ...(ii) <br> on solving (i) and (ii) we get, `prop = 45^@`. <br> <img src="//d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C05_E01_033_S01.png" width="80%">.