Question

A balloon tied up with a wooden piece is moving upward with velocity of 15 m/s. At a height of 300m above the ground, the wooden piece is detached from the balloon. How much time will it take to reach the ground?​

Answer 1

Given :

• A balloon tied up with a wooden piece is moving upward with velocity 15 m/s.

• Height = 300m.

To find :

How much time will it take to reach the ground ?

Solution :

We know that,

The wooden block is moving upward,

So, $\large \sf V_0 \ = \ -15 m/s$

Displacement,

We know that the common equation from which we can find the time is :

$\large \sf \longrightarrow S = ut + \dfrac {1}{2} at^2$

From the above equation,

Put g on the place of a because here gravity is working with respect to time.

$\large \longrightarrow \sf h = V_0t +\dfrac {gt^2}{2}$

Where,

• H = Height above the ground.
• t = Time.
• G = Gravity.

Therefore,

$\large \longrightarrow \sf 4.9 t^2 - 15 t - 300 = 0$

$\large \longrightarrow \sf D \ = \ 15^2 - 4 \times 4.9 \times (-300)$

$\large \longrightarrow \sf 225 - 4 \times 4.9 \times (-300)$

$\large \longrightarrow \sf 225 + 5880$

$\large \longrightarrow \sf 6105$

Now,

$\large \longrightarrow \sf t = \dfrac {15 + \sqrt{6105}}{2 \times 4.9}$

$\large \longrightarrow \sf t = \dfrac {15 + \sqrt{6105}}{\dfrac {49}{5}}$

$\large \longrightarrow \sf t = \dfrac {(15 + \sqrt{6105} \times 5)}{49}$

$\large \longrightarrow \sf t = \dfrac {75 + 5 \sqrt{6105}}{49}$

$\large \longrightarrow \sf t = 9.5 sec$

It will take 9.5 seconds to reach the ground.