Free Fall/Motion: Homework Help

[Grove]

Basically I'm just after confirmation that I am on the right track and that I haven't stuffed up somewhere along the lines. Thanks :)

PS. Sorry about the formatting.

Homework Statement

A mischievous student is taking a hot air balloon ride. The balloon is traveling horizontally at 9.0m/s and is 110 m above the ground. The student notices a little bunny rabbit, off into the distance, in the balloon’s direction of travel. Being mischievous, the student prepares to drop a water balloon on the poor defenseless rabbit. Rather than throw the water balloon, the student simply let's go of it over the edge.

(a) Draw a diagram of the water balloon’s trajectory. On your diagram, label the initial and final height, initial horizontal distance as well as the x and y components of the initial and final velocities. Don’t forget to specify your coordinate system, i.e. which directions are positive.
(b) Calculate the time taken for the water balloon to hit the ground.
(c) Calculate at what horizontal distance from the bunny rabbit, must the student drop the water balloon to make sure it hits the target. Label this distance on your diagram.
(d) Calculate the speed at which the water balloon hits the poor bunny.

Homework Equations

V = Vo + at

X = Xo + volt + 1/2at^2

V^2 = Vo^2 + 2a(X - Xo)

The Attempt at a Solution

a) Ignore this question since I am confident of this as it is just a diagram.

b)
Y = Yo + Voy t + 1/2at^2 (where Y = 0, Yo = 110m)
therefore;
-Yo = 1/2at^2
-110m = 1/2 x -9.8 x t^2
-110m = -4.9 x t^2
t = SQR ROOT (-110 / -4.9)
t = 4.7s

c)
d = Vox x t (since ax = 0)
d = 9.0m/s x 4.7s
d = 42.3m

d)
V^2 = Vo^2 + 2a(Y - Yo)
V = SQR ROOT (Vo^2 + 2a(Y - Yo))
V = SQR ROOT (9.0^2 + 2a(0 - 110))
V = 47.3m/s

 

[kuruman]

I didn't do the actual calculations, but the method looks fine.

 

[Grove]

Yeah the calculations themselves are all correct (double-checked) but was hoping the method and the way I did it was all correct, especially making sure I used the right vector components etc.

Actually now that I look. The final velocity value I have, is that only for the vertical component of the final velocity? If so, would I just need to add the vectors of vertical (47.3) + horizontal (9.0) to get the actual final velocity? (Answer 48.1 m/s)

Please confirm?

Thanks

 

[leena19]

Yeah the calculations themselves are all correct (double-checked) but was hoping the method and the way I did it was all correct, especially making sure I used the right vector components etc.

Actually now that I look. The final velocity value I have, is that only for the vertical component of the final velocity? If so, would I just need to add the vectors of vertical (47.3) + horizontal (9.0) to get the actual final velocity? (Answer 48.1 m/s)

Please confirm?

Thanks

d)
V^2 = Vo^2 + 2a(Y - Yo)
V = SQR ROOT (Vo^2 + 2a(Y - Yo))

I'm not very sure but if this is the equation you've used to find the vertical component of the velocity then V0 should be 0,not 9m/s which is only the horizontal component of the velocity.
Once you find this,then I guess you can find the speed at which the balloon strikes the ground by calculating the resultant velocity.

 

[Grove]

Oh I can't believe I didn't pick that up. Thanks for that.

So calculations change to 46.4m/s for vertical and horizontal stays at 9.0m/s which gives me the resultant of 47.3m/s (oddly enough).

 

[kuruman]

Oh I can't believe I didn't pick that up. Thanks for that.

So calculations change to 46.4m/s for vertical and horizontal stays at 9.0m/s which gives me the resultant of 47.3m/s (oddly enough).

It appears that, when all is said and done, you will end up with the same number in part (d). That's because

v_x² = 9.0² and
v_y² = 0²+2*9.8*110

 

[Grove]

It appears that, when all is said and done, you will end up with the same number in part (d). That's because

v_x² = 9.0² and
v_y² = 0²+2*9.8*110

That's true. Thanks guys! :)