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 17 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
To find the velocity of the water balloon when it hits the ground, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (the velocity when the balloon hits the ground)
u = initial velocity (the velocity when the balloon was pitched)
a = acceleration (the acceleration due to gravity, -9.8 m/s²)
s = displacement (the height the balloon was pitched, 17 m)
In this case, we need to find v.
First, we rewrite the equation as:
v² = u² + 2as
Since the balloon starts from rest when it is pitched, the initial velocity (u) is 0. Thus, the equation becomes:
v² = 0² + 2(-9.8)(17)
v² = 0 + (-9.8)(17)(2)
v² = -9.8(34)
v² = -333.2
To solve for v, we take the square root of both sides:
v = √(-333.2)
Keep in mind that the square root of a negative number gives an imaginary result. In this case, it means that the water balloon would not hit the ground with a positive velocity while bursting. Instead, it would fall downward under gravity until it bursts upon impact.
Therefore, the velocity of the water balloon when it hits the ground is 0 m/s.
At the top of the toss, all the KE has been converted to PE, so
1/2 mv^2 = mgh
or,
v^2 = 2gh