Question

A projectile has a maximum range of 16 km. Atthe heighest point of its motion, it explodes intotwo equal masses. One mass drops verticallydownwards. The horizontal distance covered bythe other mass from the time of explosion is:(1) 32 km (2) 24 km (3)16 km (4)8 km​

Answer 1

Answer:

The answer will be 16 km

Explanation:

According to the problem the maximum range of the projectile is given and the mass of the two explodes are said to be equal = m

Now when the mass is at the highest position the vertical component is zero and the horizontal component is ucosθ

Now we know that for maximum range θ = 45 degree

Therefore the horizontal component = ucos45 = u/√2

Now the initial momentum is u/√2

As said after the explotion it is divided into two equal masses m/2 and m/2

now one m/2 has zero and another m/2 has 2 x u/√2 = √2u

Now time to reach the ground is usin45 degree/g = u/√2g

Hence the horizontal distance covered by the other mass be √2u x  u/√2g= u^2/g

Now it is given that u^2/g = 16 km

hence the distance covered is 16 km