- #1
HumorMe81
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A firecracker with mass 100g is propelled vertically with a launch force of 1.2N applied over 5s, after which it explodes into two 50g fragments. One fragment goes horizontally at 15m/s and the other goes to the right at angle 53 degrees with the horizontal at speed 25 m/s. find the distance between them once they hit the ground.
This is what I did:
Force up - Fg = ma
1.2N - 1N = .1kg a
a = 2m/s^2 upward
I used this info to calculate the speed right before the explosion:
V = Vi + at
V = 0 + 2m/s^2 * 5s= 10m/s
The distance elapsed in this time is:
Y = Vi*t + 1/2gt^2
Y = 1/2 (2m/s^2)*(5s)^2 = 25m
Since the problem says after the 5s the cracker explodes, I thought the fragment that goes left takes 5s to come back down and the distance it would travel is 75m. But I'm wrong. It travels 33m.
X = Vi*t + 0
The fragment moving to the right is at an angle so it has a horizontal and vertical components. Vx is 15m/s and Vy is 20m/s.
Can someone guide me please? Thanks
This is what I did:
Force up - Fg = ma
1.2N - 1N = .1kg a
a = 2m/s^2 upward
I used this info to calculate the speed right before the explosion:
V = Vi + at
V = 0 + 2m/s^2 * 5s= 10m/s
The distance elapsed in this time is:
Y = Vi*t + 1/2gt^2
Y = 1/2 (2m/s^2)*(5s)^2 = 25m
Since the problem says after the 5s the cracker explodes, I thought the fragment that goes left takes 5s to come back down and the distance it would travel is 75m. But I'm wrong. It travels 33m.
X = Vi*t + 0
The fragment moving to the right is at an angle so it has a horizontal and vertical components. Vx is 15m/s and Vy is 20m/s.
Can someone guide me please? Thanks