a hot-air balloon having initial velocity vi as it ascends. Then, a water balloon is dropped from the basket when it is 135 meters high and would hit the ground after 9.00 seconds. a) find the velocity of the hot-air balloon upon release of the water balloon.
The person in the basket fears for the safety of those below and yells out a warning. The speed of sound is 343 m/s. b)assuming he can be heard from this height, how long does it take for the warning to reach the ground?
After hearing the warning, a bystander decides to take matters into his own hands and shoots the water balloon with an arrow.
c) if the arrow has an initial velocity of 50m/s straight up at what elevation does arrow strike the water balloon?
a) To find the velocity of the hot-air balloon upon release of the water balloon, we can use the equation of motion:
v = u + at
where:
v = final velocity of the hot-air balloon
u = initial velocity of the hot-air balloon
a = acceleration (assumed to be constant)
t = time
Let's assume the upward direction as positive. The acceleration of the hot-air balloon can be considered as -9.8 m/s² (due to gravity).
At the release of the water balloon, the time taken is 9.00 seconds.
The final velocity is zero (as the balloon starts to ascend).
Using the equation:
0 = vi - 9.8 * 9
Solving for vi:
vi = 9.8 * 9
vi ≈ 88.2 m/s
Therefore, the velocity of the hot-air balloon upon release of the water balloon is approximately 88.2 m/s.
b) To find the time taken for the warning to reach the ground, we will use the speed of sound formula:
v = d / t
where:
v = speed of sound (343 m/s)
d = distance traveled (total height of the hot-air balloon: 135 m)
t = time taken (unknown)
Rearranging the formula:
t = d / v
t = 135 / 343
t ≈ 0.394 seconds
Therefore, it takes approximately 0.394 seconds for the warning to reach the ground.
c) To find the elevation at which the arrow strikes the water balloon, we can use the equation of motion:
h = h0 + (v0 * t) + (0.5 * a * t²)
where:
h = final elevation
h0 = initial elevation (135 m)
v0 = initial velocity of the arrow (50 m/s)
a = acceleration (due to gravity: -9.8 m/s²)
t = time taken
We are looking for the value of h when v = 0, as the arrow reaches its highest point before striking the water balloon.
At the highest point, v = 0, so the equation becomes:
0 = 135 + (50 * t) + (0.5 * (-9.8) * t²)
Simplifying the equation:
0 = 135 + 50t - 4.9t²
Solving this quadratic equation will give us the value of t, which we can then substitute back into the equation to find the final elevation h.
However, in this case, if we consider air resistance negligible, the arrow will reach the exact same elevation (135 m) when it strikes the water balloon.
Therefore, the arrow will strike the water balloon at the same elevation as the initial height of 135 meters.
To solve these problems, we will need to do some calculations using the equations of motion.
a) To find the velocity of the hot-air balloon upon release of the water balloon, we can use the equation of motion:
vf = vi + at
Since the water balloon is dropped, its initial velocity can be assumed to be 0 m/s. The final velocity is what we want to find, and the time is given as 9.00 seconds. Acceleration, a, can be calculated using the equation:
s = vit + 0.5at^2
where s is the height (135 meters) and vi is the initial velocity of the balloon. Rearranging this equation to solve for a, we get:
a = 2s / t^2
Substituting the values, we have:
a = 2(135 m) / (9.00 s)^2
Now, we can substitute the calculated value of a into the first equation to find the final velocity, vf.
b) To calculate the time it takes for the warning to reach the ground, we need to consider the speed of sound. We have the height of the balloon from which the warning is being yelled out (135 meters). The speed of sound is given as 343 m/s. We can use the equation:
t = d / v
where t is the time, d is the distance, and v is the speed of sound. In this case, the distance is the height of the balloon.
c) To find the elevation at which the arrow strikes the water balloon, we can use the equation of motion for vertical motion:
h = vi*t - 0.5gt^2
where h is the elevation, vi is the initial velocity of the arrow (50 m/s), t is the time taken to reach the elevation, and g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearrange this equation to solve for t and then substitute the calculated value of t back in to find the elevation, h.
Using these equations and the given values, you can calculate the required answers.