You are trying to find out how high you have
to pitch a water balloon in order for it to burst
when it hits the ground. You discover that
the balloon bursts when you have pitched it
to a height of 14 m.
With what velocity did the balloon hit
the ground? The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m/s
Well, it seems like our water balloon decided to take a leap of faith from a height of 14 m! To find the velocity at which it hit the ground, we can use the kinematic equation for free-falling objects:
v² = u² + 2as
where v is the final velocity, u is the initial velocity (which is 0 since it starts from rest), a is the acceleration due to gravity (9.8 m/s²), and s is the distance traveled (14 m).
Plugging in the values, we get:
v² = 0² + 2 * 9.8 * 14
v² = 0 + 274.4
v ≈ √274.4
v ≈ 16.56 m/s
So, our water balloon hit the ground with a velocity of approximately 16.56 m/s. Keep in mind that this is a calculated value and may vary slightly in reality due to factors like air resistance or the bravery of the water balloon.
To determine the velocity at which the water balloon hits the ground, we can use the equations of motion.
First, let's define the initial height of the water balloon as h = 14 m, the final height as h' = 0 m, and the acceleration due to gravity as g = 9.8 m/s^2.
The equation we can use is:
v^2 = u^2 + 2ad
Where:
v = final velocity (unknown)
u = initial velocity (can be assumed as 0 m/s since the balloon is pitched upwards)
a = acceleration due to gravity (-9.8 m/s^2 since it acts in the opposite direction of motion, downwards)
d = distance or displacement (h' - h = -14 m)
Plugging in these values, we have:
v^2 = 0^2 + 2(-9.8)(-14)
Simplifying this equation, we get:
v^2 = 2(9.8)(14)
v^2 = 274.4
Finally, taking the square root of both sides, we find:
v = √(274.4)
v ≈ 16.56 m/s
Therefore, the velocity at which the water balloon hits the ground is approximately 16.56 m/s.
To determine the velocity with which the water balloon hits the ground, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of an object (in this case, the water balloon) remains constant as long as no external forces act on it.
At the highest point of the water balloon's trajectory (when it is pitched to a height of 14 m), all of its initial kinetic energy is converted into potential energy:
mgh = 1/2 mv²
Where:
m = mass of the water balloon (assumed constant)
g = acceleration due to gravity (9.8 m/s²)
h = height to which the balloon is pitched (14 m)
v = velocity with which the balloon hits the ground (unknown)
Simplifying the equation, we have:
gh = 1/2 v²
Plugging in the known values:
(9.8 m/s²) * (14 m) = 1/2 v²
137.2 m²/s² = 1/2 v²
Now, we can solve for v by isolating it on one side of the equation. Multiply both sides by 2, then take the square root of both sides:
2 * 137.2 m²/s² = v²
274.4 m²/s² = v²
√(274.4 m²/s²) = v
v ≈ 16.6 m/s
Therefore, the velocity with which the water balloon hits the ground is approximately 16.6 m/s.
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*14 = 274.4,
Vf = 16.6m/s.