Question

A shell is projected from a level ground with a velocity of 20m/s at 45‚àò to the horizontal. When the shell is at highest point, it breaks into two equal fragments. One of the fragment whose initial velocity after the explosion is zero, falls vertically downward. At what distance from the point of projection does the other fragment fall? (g=10ms‚àí2)

Answer 1

by momentum conservation
the velocity of other fragment is 2 times of initial horizontal velocity
= 20√2

time to reach highest point = √2 sec = time to reach ground again

so the net range of the other fragment is
20√2×√2 + 10√2×√2
= 60m

hope it helps you
@di

Answer 2

Given;-

u = 20m/s

θ = 45°

v1 = 0 m/s

To Find -

distance from the point of projection where the fragment falls

Solution;-

Horizontal velocity =  ucosθ = $\frac{20}{\sqrt{2} }$ m/s

Vertical velocity = usinθ =  $\frac{20}{\sqrt{2}}$ m/s = u₁

From the law of conservation of momentum ;-

Intial momentum = final momentum

(m1 + m2)u = m1v1 + m2v2

here, the object breaks into two equal halves

∴  m1 = m2 = m2 (let)

∴ 2(m2)u = m2 × 0 + m2 × v2

⇒ v2 =  2u = 20√2 m/s

Time taken by the fragment to reach the highest point = t

v = u₁ - gt

⇒ $0 = \frac{20}\sqrt{2}} - 10t$

⇒ $t = \sqrt{2} s$

Distance covered by the fragment from the highest point with velocity v2 = r

r = vt = √2 × 20√2 = 40m

∵ Total distance from the point of projection  = 2r = 80m