Question

7. A balloon is going upwards with velocity
12m/s. It releases a packet when it is at
height of 65m from ground. How much
time will the packet take to reach the
ground (g = 10m/s)?​

Answer 1

Given :-

Initial velocity of the balloon = 12 m/s

Final velocity of the balloon = 0 m/s

Height from the ground = 65 m

To Find :-

Time taken by the packet to reach the ground.

Analysis :-

Here we are given with the height, gravity, initial and final velocity of the balloon.

In order to find the time taken substitute the given values in the question using the second equation of motion and find the time taken by the packet to reach down accordingly.

Solution :-

We know that,

• u = Initial velocity
• a = Acceleration
• t = Time
• s = Displacement

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=ut+\dfrac{1}{2} gt^2}}$

Given that,

Displacement (s) = 65 m

Initial velocity (u) = -12 m/s

Gravity (g) = 10 m/s

Substituting their values,

$\sf 65=-12t+\dfrac{1}{2} \times 10t^2$

$\sf 65=-12t+\dfrac{10t^2}{2}$

$\sf 65=-12t+5t^2$

$\sf 5t^2 -12t-65=0$

$\sf 5t(t-5)+13(t-5)=0$

$\sf ( t - 5)(5t + 13) = 0$

$\sf t=5 \ or \ t=\dfrac{-13}{5}$

Therefore, the time taken by the packet to reach the ground is 5 seconds.