Verified answer

3 parts to this problem

h = 205 m

vx = 1.1 m/s

vy = 2.5 m/s

t2 = 10 s

1st part, figure out how far along the horizontal the balloon moved while rising to 205 m.

t1 = h/vy

sx1 = vx*t1

2nd part, figure out how far along the horizontal the balloon moved while at a constant height for 10 s.

sx2 = vx*10

3rd part, figure out how long it takes the bag to fall and then use that to figure out how far it moved.

t3 = sqrt(2*h/g) from s = s0 + v0y*t + 1/2at^2

sx3 = vx*t3

The total distance traveled is S = sx1 + sx2 + sx3

Presto!

Plug and chug.

italian baby

This is 2-D motion, so you need to break it up into 2 1-D problems. Let x be the horizontal direction and y be the vertical.

First determine where the balloon is when the sandbag is dropped:

In the x direction, v = 1.10, we want x0=vt, but we don't know t

In the y direction, v = 2.50m/s and the height attained is 205m, so the time to rise is 205/2.5. Then 205 is maintained for 10s, so the total time is 205/2.5 + 10

x0= 1.10(205/2.5 + 10)

Now we have to determine how far the sandbag lands when it is dropped. Again break it up into 2 1-D problems

In the x direction we want x1 = vt, v=1.10 again, but again we don't know t.

In the y direction we know initial vel =0 (why?) gravity is 9.8m/s/s and the displacement is 205m Note I have taken down as positive here...it doesn't matter as long as I am consistent. So we can find the time using

y = vit + 1/2at^2 or

205 = 0t + 1/2 g t^2

t = sqrt(205(2)/g)

then x1 = 1.1x sqrt(205(2)/g)

finally we add x1 and x0

hope that helps I baby!

it would take the balloon 82 seconds to reach the height of 205m (205m / 2.5m/s). so multiply 1.10 by 82 seconds +10 seconds and you get 101.2m horizontal movement. The sandbag would fall straight down. So the answer is 101.2m from the launch site