Answer:
Explanation:
Given parameters
initial velocity u = 4.9m/s
height of drop H = 100m
acceleration due to gravity g = -9.8m/s²(since the body is ascending i.e moving against gravity)
Required
time taken by the packet to reach the ground 't'
Using the equation of motion
S = ut+1/2gt²
0-100 = 4.9t+1/2(-9.8)t²
-100 = 4.9t-1/2(9.8)t²
-100 = 4.9t-4.9t²
4.9t-4.9t² +100 = 0
4.9t²-4.9t² -100 = 0
Multiplying through by 10
49t²-49t-1000 = 0
Using the general equation t = (-b±√b²-4ac)/2a
t = -(-49)±√(-49)²-4(49)(-1000)/2(49)
t = 49±√(2401+196000)/2(49)
t = 49±√(198401)/98
t = 49±√(198401)/98
t = 49±449.42/98
t = (49+449.42)/98
t = 494.42/98
t = 5.04secs
Hence, it took the packet 5.04secs to reach the ground
The final velocity is calculated using the formula v = u + at
v = u-gt
v = 4.9-(9.8)(5.04)
v = 4.9-49.392
v = -44.492m/s
This shows that the final velocity is 44.49m/s(moving downwards)
