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# Physics/University Physics problem.

Question
You drop a water balloon straight down from your dormitory window 67.0 m above your friend's head. At 1.48 s after you drop the balloon, not realizing it has water in it your friend fires a dart from a gun, which is at the same height as his head, directly upward toward the balloon with an initial velocity of 21.5 m/s.
(a) How long after you drop the balloon will the dart burst the balloon?
(b) How long after the dart hits the balloon will your friend have to move out of the way of the falling water? Assume the balloon breaks instantaneously at the touch of the dart.

Hello Andrew,

The balloon will fall with acceleration of 9.8 m/s^2. After the dart bursts the balloon, the water (most of it anyway) will continue falling with acceleration of 9.8 m/s^2 and without any momentary interruption of the increase of the velocity as the dart passed through. Let's consider down to be the positive direction. So the dart's initial speed is -21.5 m/s.

a. Let that balloon's position after 1.48 s be point 1 and the position where the balloon is burst be point 2. At t = 1.48 s, the balloon's position, and velocity can be found using the kinematic formulas
y = Vi*t + (1/2)*a*t^2
and
Vf = Vi + a*t
I'll use y1 as the distance down to point 1 and t1 and V1 as the time and velocity when at point 1. The time t1 is given: 1.48 s. The balloon's initial speed is zero, so the formulas reduce to
y1 = (1/2)*a*t1^2 and V1 = a*t1
You can work those out.

So the dart to balloon distance when the dart is fired is 67 m - y1. Let the total time that the balloon has fallen when it is burst be t2.

The closing velocity is
Balloon's velocity - dart's velocity = a*t2 - (-21.5 m/s + a*(t2-t1))
Note that gravity acted on the dart only for time t2-t1. So the closing velocity reduces to
Vclose = a*t2 + 21.5 m/s -a*t2 + a*t1 = 21.5 m/s + a*t1
where a*t1 is V1 which you calculated above. It turns out that the closing speed of the balloon and dart is constant!

We now know the distance to be covered and Vclose, so you can easily find the time to impact.

b. This is simply the time to fall 67 m. Use the 1st kinematic formula I gave you in part a with y = 67 m. Solve for t.

I hope this helps,
Steve

Physics

Volunteer

#### Steve Johnson

##### Expertise

I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

##### Experience

I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

Education/Credentials
BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University