Question

A balloon starts rising from ground from rest with an upward acceleration 2 m/s2. Just after 1 second, a stone is dropped from it. The time taken by stone to strike the ground is nearly?

Answer 1

The balloon starts to rise vertically upward with an acceleration of $2m/s^{2}$
After one second (First Case)
Therefore in this case; $u = 0 ; t =1; a=2$
The distance traveled in this case by the formula $s=ut+ \frac{1}{2}at^{2}$
∴ s = 1m
Case Two:-
From there a stone was dropped...
Therefore the stone was falling under the effect of gravity
Therefore in this case; $u=0;s=1;a=9.8$
The time can be calculated in this case by the formula $s=ut+ \frac{1}{2}at^{2}$
∴ t ≈ 0.45s

Answer 2

u=0 , a=2 and t=1

v=u+at.

v = 0+2×1

v = 2 .......... 1.

v^2=u^2+2as

2^2 = 0+2×2×s

4=4s

s=1 m ........... 2.

Time taken by stone to strike the ground is

s = ut+1/2gt^2

1 = -2×t+1/2×9.8×t^2

1 = -2t+4.9t^2

4.9t^2-2t-1 {Quadratic equation

ax^2 + bx + c}

Acc. To this formula

D=b^2-4ac

x=-b±√D/2a

t = -(-2)±√2^2 - 4×4.9×(-1 )/ 2×4.9

t = 2±√4+19.6 / 9.8

t= 2±√23.6 / 9.8

t= 2+4.8 / 9.8

t=0.7 (approximate)