Question

A balloon is ascending at the rate of 10m/s when it is at a height of 75m from the ground a packet is dropped from it. The packet will reach the ground after time of?

Answer 1

Answer:

the answer is 5 seconds

Answer 2

Answer:

The packet will reach the ground after a time of 3 sec.

Explanation:

From the third equation of motion we have,

$v^2-u^2=2gs$      (1)

Where,

v=final velocity of the body

u=initial velocity of the body

a=acceleration of the body

s=distance traveled by the body

g=acceleration due to gravity=10m/s²

From the question we have,

u=10m/s

s=75m

By substituting the values in equation (1) we get;

$v^2-(10)^2=2\times 10\times 75=1500$

$v^2=1500+100=1600$

$v^2=1600$

$v=\sqrt{1600}=40m/s$         (2)

And from the first equation of motion, we have,

$v=u+gt$              (3)

By putting the values in equation (3) we get;

$40 =10+10\times t$

$t=\frac{40 -10}{10}=\frac{30}{10}=3sec$

Hence, the packet will reach the ground after a time of 3 sec.