please help me with this question
Answers
Answer:refer to attachment please. And i am sorry i cant answer second part.
Answer:
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