Question

A balloon is ascending at the rate of 9.8ms^(-1) at a height of 39.2m above the ground when a food packet is dropped from the balloon. After how much time and with what velocity does it reach the ground? Take g=9.8ms^(-2) .​

Answer 1

Answer:

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

Explanation:

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Answer 2

DATA

S=39.2m

U= -9.8m/s

(as balloon was ascending with same speed but packet

Direction is now opposite)

Formula

a= 9.8m/s2

$S =ut + \frac{1}{2} at^{2}$

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)

$0=t ^{2} - 2t - 8$ (factorise)

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

thus because t-4=0 ,t+2=0 now ,t=4 ,t=-2

time cannot be negative so

t=4s

v=u+at

=-9.8 + 9.8×4

=29.4m/s