SOLUTION: A balloon rising vertically with a velocity of 16 ft/s releases a sandbag. At an instant when the balloon is
64 ft above the ground.
(a) Compute the position and velocity of the
Question 1177201: A balloon rising vertically with a velocity of 16 ft/s releases a sandbag. At an instant when the balloon is
64 ft above the ground.
(a) Compute the position and velocity of the sandbag at the following times after its release: ¼ sec, ½
sec, 1 sec, 2 sec.
(b) How may seconds after its release will the bag strike the ground?
(c) With what velocity will it strike? Found 2 solutions by Solver92311, ikleyn:Answer by Solver92311(821) (Show Source):
The height in feet as a function of time in seconds for a falling object that has an initial height of and assuming no initial velocity, near the surface of planet Earth is:
So for this situation:
Evaluate , , , and . Then set the function equal to zero and solve for . The instantaneous velocity is given by the first derivative of the height function.
Evaluate the instantaneous velocity function at the calculated ground impact time.
John
My calculator said it, I believe it, that settles it
From
I > Ø
There is one hidden feature in this problem.
This feature is the initial speed of the sandbag at the time moment of releasing.
This initial speed is 16 ft per second, directed vertically up at the release moment.
Taking it into account, the equation for the position / (the height) of the sandbag as a function of the time is
h(t) = -16t^2 + 16t + 64.
Having it established, you have a standard problem on a projectile launched vertically up with the given velocity.
After that, you can easily solve the problem and answer all the questions on your own.
Consider these lessons as your textbook, handbook, tutorials and (free of charge) home teacher.
Read them attentively and learn how to solve this type of problems once and for all.
