A stunt pilot is attempting to drop a water balloon from a moving airplane onto a target on the ground. The plane moves at a speed of 77.0 m/s and a 39° above the horizontal when the balloon is released. At the point of release, the plane is at an altitude of 500 m.
(a) How far horizontally, in meters and measured from a point directly below the plane's initial position, will the balloon travel before striking the ground?
(b) At the point just before the balloon strikes the ground, what angle does its velocity make with the horizontal? Give your answer as an angle measured in degrees below the horizontal.
(a) the horizontal speed is 77.0 cos39° = 59.84 m/s
the height t seconds after release, is
h(t) = 500 + 77.0 sin39° t - 4.9t^2
h=0 at t=16.19 s
so the balloon travels 59.84 * 16.19 = 968.81 m
see what you can do with (b)
To solve this problem, we need to break it down into two parts:
Part 1: Finding the time of flight of the balloon.
First, let's find the time it takes for the balloon to hit the ground. We will assume there is no air resistance.
We have the initial vertical velocity (Vy = 0 m/s) and the acceleration due to gravity (g = 9.8 m/s^2). We can use the kinematic equation:
h = Vit + (1/2)gt^2,
where h is the altitude of the plane (500 m), Vi is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight.
Since Vi is 0 m/s, the equation simplifies to:
h = (1/2)gt^2.
Plugging in the given values:
500 = (1/2)(9.8)t^2
Solving for t:
t^2 = (2*500)/9.8
t^2 = 102.04
t = √102.04
t ≈ 10.102 s
So, the time of flight for the balloon is approximately 10.102 seconds.
Part 2: Finding the horizontal distance traveled by the balloon.
Now that we have the time of flight, we can find the horizontal distance traveled by the balloon. We'll use the equation:
d = Vx * t,
where d is the horizontal distance, Vx is the horizontal component of the velocity, and t is the time of flight.
To find Vx, we can use trigonometry:
Vx = V * cos(θ),
where V is the velocity of the plane (77.0 m/s) and θ is the angle above the horizontal (39°).
Plugging in the values:
Vx = 77.0 * cos(39°),
Vx = 77.0 * 0.766 = 59.202 m/s.
Now, we can find the horizontal distance traveled:
d = 59.202 * 10.102,
d ≈ 597.92 m.
So, the balloon will travel approximately 597.92 meters horizontally before striking the ground.
Part 3: Finding the angle of velocity just before impact.
At the point just before the balloon strikes the ground, the horizontal component of its velocity will remain the same (Vx ≈ 59.202 m/s). However, the vertical component will be opposite in direction due to gravity, resulting in a downward angle.
To find this angle, we can calculate the tangent of the angle:
tan(θ) = Vy / Vx,
where θ is the angle we're looking for.
Plugging in the values:
θ = tan^(-1)(Vy / Vx),
θ = tan^(-1)((-9.8) / 59.202),
θ ≈ -9.47°.
So, the angle of velocity just before the balloon strikes the ground is approximately 9.47° below the horizontal.
To solve this problem, we can break it down into two parts: the horizontal motion and the vertical motion of the water balloon.
(a) To find how far horizontally the balloon will travel before striking the ground, we need to find the time it takes to reach the ground and then multiply it by the horizontal velocity. The horizontal motion does not affect the balloon's vertical motion, so we can consider them separately.
First, let's find the time it takes for the balloon to reach the ground. We can use the vertical motion equation:
y = y0 + V0y * t + (1/2) * a * t^2
where:
y = final position in the y-direction (0 since the balloon will hit the ground)
y0 = initial position in the y-direction (500 m)
V0y = initial vertical velocity (V0y = V0 * sin(θ) where V0 = 77.0 m/s and θ = 39°)
a = acceleration in the y-direction (a = -9.8 m/s^2, assuming downward as negative)
t = time
Plugging in the given values, we get:
0 = 500 + (77.0 * sin(39°)) * t + (1/2) * (-9.8) * t^2
Simplifying this equation gives us a quadratic equation:
-4.9 * t^2 + 38.5 * t + 500 = 0
Now we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
where:
a = -4.9
b = 38.5
c = 500
Using the quadratic formula, we find two solutions: t = 5.56 s and t = 18.13 s. Since we're only interested in the positive value for time (since time cannot be negative), the water balloon takes approximately 5.56 seconds to reach the ground.
Next, we can find the horizontal distance traveled by the balloon before striking the ground. We can use the horizontal motion equation:
x = x0 + V0x * t
where:
x = horizontal distance traveled (what we're trying to find)
x0 = initial position in the x-direction (0 since the balloon is dropped directly below the plane's initial position)
V0x = initial horizontal velocity (V0x = V0 * cos(θ) where V0 = 77.0 m/s and θ = 39°)
t = time (5.56 s)
Plugging in the given values, we get:
x = 0 + (77.0 * cos(39°)) * 5.56
Simplifying this equation gives us the horizontal distance traveled:
x ≈ 382.8 meters
Therefore, the balloon will travel approximately 382.8 meters horizontally before striking the ground.
(b) Just before the balloon strikes the ground, its vertical velocity will be its final velocity. We can use the vertical motion equation again to find the final vertical velocity:
Vfy = V0y + a * t
where:
Vfy = final vertical velocity
V0y = initial vertical velocity (V0y = V0 * sin(θ) where V0 = 77.0 m/s and θ = 39°)
a = acceleration in the y-direction (a = -9.8 m/s^2, assuming downward as negative)
t = time (5.56 s)
Plugging in the given values, we get:
Vfy = (77.0 * sin(39°)) + (-9.8) * 5.56
Simplifying this equation gives us the final vertical velocity:
Vfy ≈ -50.0 m/s (note: negative sign indicates downward direction)
Next, we can find the angle the velocity makes with the horizontal. We can use the tangent function to find this angle:
tan(θ) = Vfy / Vfx
where:
Vfx = horizontal velocity (Vfx = V0 * cos(θ) where V0 = 77.0 m/s and θ = 39°)
Plugging in the given values, we get:
tan(θ) = (-50.0) / (77.0 * cos(39°))
Simplifying this equation gives us the angle θ:
θ ≈ -39.2°
Therefore, at the point just before the balloon strikes the ground, its velocity makes an angle of approximately 39.2° below the horizontal.