Question

A balloon is ascending at the rate of 9.8m/s at a height of 39.2m above the ground,when a food packet is dropped from the balloon.After how much time and with velocity does it reach the ground. g=9.8m/s2

Answer 1

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

Answer 2

Answer:t=4 secs and v=-29.4

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