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1 answer:
Answer:
Explanation:
Initially the packet was ascending up with the balloon.
Taking upward as positive direction;
initial velocity, u = 4.9 m/s
final velocity = v m/s
initial height, h₁ = 100 m
final height, h₂ = 0
a = -9.8 m/s²
time taken = t seconds
Considering the second equation of motion.
S = ut + 1/2at^2
s = ut + 0.5at²
(h₂-h₁) = ut + 0.5at²
0- 100 = 4.9t + 0.5×(-9.8)×t²
-100 = 4.9t - 4.9t²
4.9t² -4.9t - 100 =0
t² -t - 20.41 = 0
Using : b±√(b²-4ac))/(2a).... 1
Where a =1, b = -1 c =-20.408
Substituting the values into equation 1
Solving it, we get t = 5.045s
v = u + at = 4.9 -9.8×5.045 = 4.9 - 49.44 = -44.54 m/s
The final velocity is -44.54 m/s, and it took 5.045s to reach the ground.
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Answer:
u= 20.09 m/s
Explanation:
Given that
m = 0.02 kg
M= 2 kg
h= 0.2 m
Lets take initial speed of bullet = u m/s
The final speed of the system will be zero.
From energy conservation
1/2 m u²+ 0 = 0+ (m+M) g h
m u²=2 (m+M) g h
By putting the values
0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)
u= 20.09 m/s
