Question

. A balloon is ascending with a velocity of 9.8

−1

).; a sand bag is dropped from the balloon

when it is at a height of 39.2m from the ground. Calculate the time taken by bag to reach the

ground.(t = 4 s​

Answer 1

Answer:

USE EQUATIONS OF MOTION!

S=39.2m , u= -9.8m/s(as balloon was ascending with same speed but packet

Direction is now opposite)

a= 9.8m/s2

S =ut + 1/2 at sq.

Substitute values and solve.(Tips:take common 9.8 on left. Cancel common factor 9.8 on both sides. Take all values to left side. Multiply by 2 on both sides)

0=t2- 2t - 8 (factorise)

( t-4) (t+2)=0

t=4s

v=u+at

=-9.8 + 9.8×4

=29.4m/s

                           OR THIS WAY AS WELL IF IT IS EASIER FOR YOU

Explanation:s=ut+1/2at2

-39.2=9.8t-4.9t2

-8=2t-t2

t2-2t-8=0

t2-4t+2t-8=0

t (t-4)2 (t-4)

t+2 t-4

t=-2 t=-4