Question

A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1 = 2.63s after the explosion. When does the second fragment reach the ground?

Answer 1

answer:

When does the second fragment reach the ground after the explosion= 6.33444 seconds is the answer

Explanation:

Initial velocity of toal mass = 40 m/s

distance travelled before reaching peak of its trajectory = (u^2)/ (2*g) = 40^2 / (2*9.8) = 81.63265 m

At the peak of its trajectory, velocity = 0

M = m1+m2

M is total mass

Let m1 fall after t1 = 2.63 s

then if initial velocity of mass be v1

the mass m1 travels down a distance of 81.63265 m from the top.

so , S = v1*t1 + (0.5*g*t1^2)

81.63265 = v1*2.63 + ( 0.5*9.8*2.63^2)

v1 = 18.152 m/s

so, by conservation of momentum

M*V = m1 *v1 + m2*v2

we have V=0 when it explodes

so 0 = m1*18.152 + m2*v2

and we have M = m1+m2

m1 = m2 as they are equal fragments

so, 0 = m1*18.152 + m1*v2

so v2 = -18.152 m/s

it means that it is travelling upward with a velocity of 18.152 m/s

the time required for the particle m2 to reach it's peak = v2/g = 18.152 /9.8 = 1.85224 s

the distance travelled from the inital point to it's peak for m2 = (v2^2)/ (2*g) = 18.152^2 / (2*9.8)= 16.81097 m

The height from the ground at it's peak for m2 = 16.81097 + 81.63265 = 98.44362 m

so time required to come down from a height of 98.44362 m = sqrt ( 2*S/g) = sqrt(2*98.44362/9.8) = 4.48224 seconds

Total time required for particle 2 = 4.48224 + 1.85224 = 6.33444 seconds

When does the second fragment reach the ground after the explosion= 6.33444 seconds is the answer