A weather balloon is floating at a constant height above Earth when it releases a pack of instruments. If the pack hits the ground with a velocity of -68.5 m/s, how far does the pack fall?

a = -9.80 m/s^2
Vf = -68.5 m/s
d = ?

vf^2=vi^2+2ad where a= 9.8

solve for d

d = Vf^2/2a

d = -68.5/(2 x -9.80)
d = 3.495 and this answer is being marked incorrect. Am i doing something wrong.

To find the distance the pack falls, we can use the kinematic equation:

d = Vf^2 - Vi^2 / (2a)

Where:
d is the distance
Vf is the final velocity (-68.5 m/s)
Vi is the initial velocity (0 m/s, since the pack was released from rest)
a is the acceleration due to gravity (-9.80 m/s^2)

Plugging in the values, we get:

d = (-68.5 m/s)^2 - (0 m/s)^2 / (2 * -9.80 m/s^2)

Simplifying the equation, we have:

d = 4692.25 m^2/s^2 / (-19.60 m/s^2)

d = -239.21 m^2/s^2

Since distance cannot be negative, we take the magnitude of the result:

d = 239.21 m

Therefore, the pack falls approximately 239.21 meters.

To find the distance (d) that the pack falls, we can use the formula of motion:

d = (Vf^2 - Vi^2) / (2a)

Where:
- Vf is the final velocity of the pack (which is -68.5 m/s in this case).
- Vi is the initial velocity of the pack when it was released from the weather balloon (since it was floating at a constant height, Vi = 0 m/s in this case).
- a is the acceleration due to gravity, which is approximately -9.80 m/s^2.

Substituting the given values into the formula, we get:

d = (-68.5^2 - 0^2) / (2 * -9.80)

Calculating this expression will give us the distance (d) that the pack falls.