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Grove
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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.
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.
V = Vo + at
X = Xo + Vot + 1/2at^2
V^2 = Vo^2 + 2a(X - Xo)
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
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 + Vot + 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