Question

A shell is launched at angle 62° above the horizontal with initial speed 30 m/s. It follows a typical projectile-motion trajectory, but at the top of the trajectory, it explodes into two pieces of equal mass. One fragment has speed 0 m/s immediately after the explosion, and falls to the ground. How far from the launch-point does the other fragment land, assuming level terrain and negligible air resistance?

Answer 1

at the top of the trajectory , the shell has only horizontal velocity component just before explosion which is given as

v_ox = initial velocity in horizontal direction just before explosion = 30 Cos62 = 14.08 m/s

M = mass of the shell just before explosion = m

m₁ = mass of the one piece after explsoion which falls to ground = m/2

m₂ = mass of the other piece after explsoion = m/2

v_1x = velocity of one piece that falls to ground = 0 m/s

v_2x = velocity of other piece= ?

using conservation of momentum along X-direction

M v_ox = m₁ v_1x + m₂ v_2x

m (14.08) = (m/2) (0) + (m/2) v_2x

v_2x = 28.2 m/s

t = time of travel for the other piece = half the time of flight for shell = v_ox Sin62/g = 30 Sin62/9.8 = 2.7 sec

X' = horizontal distance of the point where the explosion take place = (0.5 ) (range of shell) = (0.5) (v_ox² Sin2(62) /g) = (0.5) ((30)² Sin124 /9.8) = 38.07 m

D = distance from the launch point = X' + v_2x t

D = 38.07 + (28.2 x 2.7)

D = 114.21 m