Answer:
a) The horizontal velocity of the balloon just before it hits the ground is 6 m/s
b) The magnitude of the vertical velocity of the balloon just before it hits the ground is 98 m/s.
Explanation:
Hi there!
The velocity and position vectors of the water balloon are given by the following equations:
r =(x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v =(v0x, v0y + g · t)
where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive) .
v = velocity vector at time t.
a) Please, see the attached figure for a graphic description of the problem.
Considering the origin of the frame of reference as the point of launch, notice that the position vector when the balloon hits the ground is
r1 = (60, -50) m
Then:
r1x = 60 m = v0x · t
r1y = -50 m = 1/2 · (-9.8 m/s²) · t²
(notice that the initial vertical velocity is zero, see figure).
Solving r1y for t:
(-50 m · 2) / -9.8 m/s² = t²
t = 10 s
Now, let´s replace t in the r1x equation and solve it for the horizontal component of the velocity:
60 m = v0x · 10 s
v0x = 60 m / 10 s
v0x = 6 m/s
The initial horizontal component of the velocity is 6 m/s. This velocity is constant because there is no air resistance. Then, just before the balloon hits the ground, it will have a horizontal velocity of 6 m/s.
b) To calculate the vertical component of the velocity when the balloon hits the ground, let´s use the equation of the vertical component of the velocity:
v1y = v0y + g · t
Since v0y = 0
v1y = -9.8 m/s² · (10 s) = -98 m/s
The magnitude of the vertical velocity of the balloon when it hits the ground is 98 m/s.
