Question

A ballon is ascending at the rate of 14m/s at a height of 98m above the ground zwhen a packet is dropped from bollon .After how much time and with .what velocity does it reach the ground​

Answer 1

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Keywords of ballon :-

Initial velocity = 14m/s

Height reached = -98m. (since against gravity)

Acceleration due to gravity = -9.8m/s^2 (since vertical)

Now according to second equation of motion :

$s = ut + \frac{1}{2} \times a {t}^{2}$

In this equation :

• s = displacement
• u = Initial velocity
• t = time taken
• a = acceleration

Now after inputting the values in this equation we get:

$98 = 14t + \frac{1}{2} \times 9.8 {t}^{2}$

After solving we get :

t = 6s.

Thus the packet will reach ground in 6s.

Now according to first equation of motion :

$v = u + at$

After inputting the known values in this equation we get :

$v \: = 14 + ( - 9.8(6) \: )$

$v = 14 - 58.8$

v = -44.8

Thus the velocity with which it will return the ground is -44.8m/s.

Here (-) represents the downward direction of motion.

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BY √\__________ SCIVIBHANSHU

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