5. A balloon is ascending at the rate of 14 ms
height of 98 m above the ground when the food
packet is dropped from the balloon. After how
much time and with what velocity does it reach the
ground ? Take g = 9.8 ms s-2
Answers
Answer:=
u t + ½ a t²,
where,
u = initial velocity of the food packet = +14 m/s ( + sign because the velocity is directed upwards)
a = acceleration due to gravity = - 9.8 m/s² (- ve sign as the acceleration due to gravity is directed downwards)
s = displacement of the food packet = - 98 m ( -ve sign as ground is downwards from the drop point).
Substituting various values we get,
- 98 = 14 t - ½ × 9.8 × t²
=> - 98 = 14 t - 4.9 t²
=> 4.9 t² -14 t - 98 = 0
Dividing by 0.7 throughout we get
7 t² - 20 t - 140 = 0
Solving for t we get,
t = [( 20 ± 65.73)/14] = - 3.266 s or 6.12 s.
Time t = - 3.26 s is unphysical. It means that the packet reached the ground before being dropped.
So the time t taken by the food packet to reach the ground = 6.12 second after being dropped.
v = u + a t = 14 - 9.8 × 6.12 = -45.98 m/s.
The minus sign with v mean that the velocity of the food packet is directed downwards.
Explanation:
Hope it will help you