Question

A hot-air balloon rises from ground level at a constant velocity of 3.0 m/s. One minute after liftoff, a sandbag is dropped accidentally from the balloon. Calculate (a) the time it takes for the sandbag to reach the ground and (b) the velocity of the sandbag when it hits the ground.

Answer 1

Answer:

a) It takes 6,37 s b) The Velocity is -59,43 m/s

Explanation:

The initial variables of the balloon are:

Xo = 0 m

Vo = 3 m/s

After one minute the situation is the following:

t= 60 s

X1 = Xo + Vo*t

X1= 0 m + 3 m/s * 60s

X1= 180m

So when the bag falls, its initial variables are the following:

Xo = 180m

X1 = 0m

Vo = 3 m/s

V1= ?

a= -9,8 m/s2

The ecuation of movement for this situation is:

X = Vo*t + 1/2 a*$t^{2}$

So:

-180m = 3m/s*t+ 1/2*-9,8 m/s2 * $t^{2}$

To solve this we have

a=-9,8/2

b=3

c=180

The formula is:$(-b +/- \sqrt{b^{2} -4ac}) /2a$

Replacing, we get to 2 solutions, where only the positive one is valid because we are talking about time.

So the answer a) is t= 6,37 s

With that answer we can find the question b), with the following movement formula.

Vf = Vo + at

Vf = +3 m/s + (-) 9,8 m/s2 *6,37s

b) Vf = -59,43 m/s