. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At t
he same instant the balloon is released; a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon
2 answers:
Answer:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
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Answer: hello some part of your question is missing attached below is the missing detail
answer :
<em>w</em>f = M( v cos∅ )D / I
Explanation:
The Angular speed <em>wf </em>of the system after collision in terms of the system parameters and I can be expressed as
considering angular momentum conservation
Li = Lf
M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) <em>w</em>f
where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )
In terms of system parameters and I
<em>w</em>f = M( v cos∅ )D / I
Answer:
The balloon hit the ground with velocity -15.34 m/s
Explanation:
<em>Lets explain how to solve the problem</em>
You found that the best height to pitch a water balloon in order for it to
burst when it hits the ground is 12 meters.
We consider that the 12 meters is the maximum height, so the velocity
at this height is zero.
To find the velocity when the balloon hits the ground lets use the rule
<em>v² = u² + 2gh</em>, where v is the final velocity, u is the initial velocity, g is
the acceleration of gravity and h is the height.
u = 0 , h = 12 m , g = 9.8 m/s²
<em>Substitute these values in the equation above</em>
v² = 0 + 2(9.8)(12)
v² = 235.2
<em>Take square root for both sides</em>
v = ±
The velocity is downward, then it's a negative value
Then v = -15.34 m/s
<em>The balloon hit the ground with velocity -15.34 m/s</em>