Question

A balloon is launched straight upward and has a hang time of 14 s. (wow)
A.) What was the launching velocity of the balloon?
B.) How high did the balloon travel?
C.) How far did the balloon travel between t = 3 and t = 4 s?
please explain and show yr work .... show how u got yr answer

Answer 1

Wait, is all that we are given that the balloon has a hangtime of 14 seconds?

Answer 2

i agree do we get any more information to help?

Answer 3

A) at the top of the trajectory (halfway through the entire trajectory, at t = 7 s), vf = 0
using the kinematics equation:

vf = v0 + at

plug in vf = 0, a = -9.81 m/s/s and t = 7s, then solve for v0

B) kinematics equation: vf^2 = v0^2 + 2ad
again, considering the halfway point of the trajectory (at the maximum height of the trajectory)

vf = 0, v0 = whatever you got from part A, a = -9.81 m/s/s, plug in and solve for d (the height of the trajectory)

C) kinematics equation: d = v0 * t + (1/2)at^2
find the difference in d between t = 3 and t = 4

Answer 4

vf deos that stand for final velocity? and thank youuu

Answer 5

ah, yes, vf = final velocity and v0 = initial velocity

Answer 6

okayy thank youuu very much