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.
