Question

Suppose a 13.0 kg fireworks shell is shot into the air with an initial velocity of 62.0 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? B) Calculate the velocity of the 4.00 kg piece just after the separation. c)At what horizontal distance from the starting point does the 4.00 kg piece hit the ground?

Answer 1

V_o = initial velocity of launch = 62 m/s

= angle of launch = 80

the maximum height reached by the firework is given as

H = Vo² Sin² / 2g

H = (62)² (Sin80)² / (2 x 9.8)

H = 190.21 m

time taken to reach the maximum height :

t = V_o Sin80 / 9.8 = 62 Sin80 / 9.8 = 6.23 sec

horizontal distance travelled by the firework = R = V_o Cos80 t = 62 Cos80 (6.23) = 67.1 m

a)

horizontal distance at which 9 kg mass hits = 67.1 m

b)

at the topmost point :

velocity of firework in vertical direction before explosion = V_oy = 0

velocity of firework in horizontal direction before explosion =V_ox = V_o cos80= 62 cos80 = 10.8 m/s

m₁ = mass of first piece = 9 kg

m₂ = mass of second piece = 4 kg

V_1y = velocity of mass m₁ in Y-direction after explosion = 0

V_2y = velocity of mass m₂ in Y-direction after explosion

V_1x = velocity of mass m₁ in X-direction after explosion = 0

V_2x = velocity of mass m₂ in X-direction after explosion

Using conservation of momentum along Y-direction :

M V_ox = m₁ V_1x + m₂ V_2x

13 x 10.8 = 9 x 0 + 4 V_2x

V_2x = 35.1 m/s

velocity of 4 kg piece = 35.1 m/s

c)

horizontal distance travelled by 4 kg piece = 35.1 x t = 35.1 x 6.23 = 218.7 m

total distance from the initial starting point = 67.1 + 218.7 = 285.8 m