Suppose a 13.0 kg fireworks shell is shot into the air with an initial velocity

Question

Suppose a 13.0 kg fireworks shell is shot into the air with an initial velocity of 64.0 m/s at an angle of 80.0° above the horizontal. At the highest point of its trajectory, a small explosive charge separates it into two pieces, neither of which ignite (two duds). One 9.00 kg piece falls straight down, having zero velocity just after the explosion. Neglect air resistance (a poor approximation, but do it anyway).

(a) At what horizontal distance from the starting point does the 9.00 kg piece hit the ground? __________m       not 6.43

(b) Calculate the velocity of the 4.00 kg piece just after the separation. __________m/s    not 22.2

(c) At what horizontal distance from the starting point does the 4.00 kg piece hit the ground? ___________m        not 214.12

Explanation / Answer

Here, horizontal component of launch = (cos 80 x 64) = 11.11 m/sec.
Vertical initial component = (sin 80 x 64) = 63.03 m/sec.

Height attained = (v^2/2g) = (63.03^2/19.6) = 202.69 metres.

Time to max. height = (v/g) = (63.03/9.8) = 6.43 secs.

(a) Horizontal distance at time of explosion = (6.43 x 11.11) = 71.44 metres.

(b) Now at the heighest point of trajectory:

Applying conservation of momentum-

Initial momentum, Pi = 13*11.11

Final momentum, Pf = 0+4*v

Equating:

13*11.11 = 0+4*v

=> v = 36.11 m/s

(c) Since the height attained by the sheel = 202.69 m

So, time taken by it to reach the ground-

202.69 = 0+0.5*9.8*t^2

=> t^2 = 41.36

=> t = 6.43 s

So the horizontal distance of 4.0 kg piece from the starting point = 71.44+6.43*36.11 = 303.63 m

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.