Question

A balloon is ascending at the rate of 5m/s at a height of 100m above the ground when a packet is dropped from the balloon. After how much time does it reach the ground? (Take g=10m/s^2)

Answer with steps

Answer 1

Hope this could be helpful. ◉‿◉

Answer 2

Answer :-

5s

Explanation :-

Given :

• Initial velocity, u = 5m/s

• Acceleration, a = -10 m/s^2

• Displacement of the balloon when it reaches the ground is, s = -100m

To Find :

Time taken,t => ?

Solution :

From second law of motion,we get

$\sf{}\boxed{\bf{s=ut+\dfrac{1}{2}at^2}}$

$\sf{}:\implies -100=5t-\dfrac{1}{2}\times 10\times t^2$

$\sf{}:\implies -100=5t-5t^2$

$\sf{}:\implies -100-(5t-5t^2)=0$

$\sf{}:\implies -100-5t+5t^2=0$

$\sf{}:\implies 5(-20-t+t^2)=0$

$\sf{}:\implies 5\times (-20-t+t^2)=0$

$\sf{}:\implies \dfrac{5\times (-20-t+t^2)}{5}=0\times \dfrac{1}{5}$

$\sf{}:\implies (-20-t+t^2)=0$

$\sf{}:\implies t^2-t-20=0$

$\sf{}:\implies t^2+4t-5t-20=0$

$\sf{}:\implies (t^2+4t)-(5t+20)=0$

$\sf{}:\implies (t-5)(t+4)=0$

t - 5 = 0

t = 5

t + 4 =0

t = -4

Therefore,we get two values of t.

First is 5 and second is -4

Time is always positive,so we will not take -4.

Therefore, after 5s the balloon will reach the ground.