Question

Two towers AB and CD are situated at a distance 'd' apart. AB is 20m high and CD is 30m. An object of mass m is thrown from AB horizontally at 10m/s. At the same time, an object of mass 2m is thrown from top CD at an angle 60degree with the horizontal towards AB with the same magnitude of initial velocity as first. The objects collide in mid air & stick to each other.
i) Calulate the distance betweem the towers. ii)Find the position where the objects meet.​

Answer 1

Answer:

Let  the two objects meet after time t.

according to the equation of motion for AB:

x=10t  

y=1/2gt^2 =d (AB)

according to the equation of motion for :CD

x=10cos60degreet and d(CD)=10sin60degreet+1/2gt^2

when the two particles strike each other

d(AB)+10=d(CD )

1/2gt^2+10=10sin60degreet+1/2gt^2

=>10t=20/(3)^1/2m =distance travelled by particle projected from AB and t=2/(3)1/2

Let the distance =b

therefore,b=10(3)^1/2m.

Answer 2

Answer:

d = 10√3 m

40/3 m above ground

Explanation:

An object of mass m is thrown from AB horizontally at 10m/s.

AB is 20m high

so Object will always be aat height less than 20 m  from ground  to be in the air

Time to reach at ground

20 = (1/2)gT²  => T² = 4 => T = 2 Sec

in 0 to 2 sec Both objects should collide

Distance Covered in T Sec =  (1/2)*10*T² = 5T²

Other object thrown at angle 60 deg with 10 m/s

vertical Velocity = 10 Sin60 = 5√3

Distance Covered in T Sec = 5√3T + (1/2)*10T²  =  5√3T + 5T²

5√3T + 5T² = 10 + 5T²  ( as CD is 10 m more height then AB)

=> T = 2/√3

5T² = 5(2/√3)² = 20/3     (20 - 20/3)  =  40/3 m above Ground

Horizontal Covered = 10 * 2/√3  =  20/√3

Horizontal Covered = 5*2/√3  = 10/√3

Total Horizontal Distance = 20/√3 + 10/√3  = 30/√3  = 10√3