Question

Suppose a 11.0 kg fireworks shell is shot into the air with an initial velocity of 68.0 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 (in m) from the starting point does the 9.00 kg piece hit the ground? m (b) Calculate the velocity (in m/s) of the 2.00 kg piece just after the separation. m/s (c) At what horizontal distance (in m) from the starting point does the 2.00 kg piece hit the ground? m

Answer 1

bLets calculate the velocity components

Horizontal component of launch = (cos 80 x 68) = 11.8 m/sec.
Vertical initial component = (sin 80 x 68) = 66.96 m/sec.
Height attained = (v^2/2g) = (66.96^2/19.6) = 228.8 metres.
Time to max. height = (v/g) = (66.96/9.8) = 6.83 secs.

a) Horizontal distance at time of explosion = (6.83 x 11.8) = 80.62metres.
Momentum = (mv) = 11 x 11.8) = 129.8 kg/m/sec.
The 9kg. part flies backwards at the forward velocity of 11.8 m/sec., cancelling its forward speed. It is stationary, and drops vertically down.
(9 x 11.8) = 106.2 kg/m/sec. momentum.

For 2 kg piece,   momentum = (129.8-106.2) = 23.6

vELOCITY OF 2 KG PIECE = 23.6 +11.8 = 35.4 m/s ANSWER FOR (B)

a)
Its time to ground = sqrt.(2h/g) = sqrt.((228.8 x 2)/9.8) = 6.83 secs.
Distance gained = (6.83 x 35.4) = 241.8 metres.

c)

Total distance = (241.8 + 80.62) = 322.5 metres