Question

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 83.4 m/s and 41degrees above horizontal when the balloon is released. At the point of release, the plane is at an altitude of 620m. a. How far horizontally, in meters and measure 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 and the ground, what angle does it’s velocity make with the horizontal? Give your answer as an angle measured in degrees below the horizontal.

Answer 1

To solve this problem, we can break it down into two parts: finding the horizontal distance traveled by the balloon before striking the ground and determining the angle of its velocity just before impact.

a. To find the horizontal distance traveled by the balloon, we can use the horizontal component of the velocity and the time it takes to reach the ground.

The initial horizontal velocity (Vx) can be calculated using trigonometry:
Vx = V * cos(θ)

where V is the speed of the plane (83.4 m/s) and θ is the angle above the horizontal (41 degrees).

Vx = 83.4 m/s * cos(41°)
Vx ≈ 63.957 m/s

To calculate the time taken for the balloon to reach the ground, we can use the vertical motion equation:

h = Vyi * t + (1/2) * g * t^2

where h is the initial altitude (620 m), Vyi is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken to reach the ground.

Since the balloon is released with no initial vertical velocity (Vyi = 0), the equation simplifies to:

h = (1/2) * g * t^2

Solving for t:

t^2 = (2 * h) / g
t = sqrt((2 * 620 m) / 9.8 m/s^2)
t ≈ 8.05 s

Now, we can calculate the horizontal distance (d) using the time and horizontal velocity:

d = Vx * t
d = 63.957 m/s * 8.05 s
d ≈ 515.278 m

Therefore, the balloon will travel approximately 515.278 meters horizontally before striking the ground.

b. To determine the angle of the velocity just before impact, we can use trigonometry. The vertical component of the velocity just before impact (Vyf) can be found using the vertical motion equation:

Vyf = Vyi + g * t

Since the balloon falls from rest vertically (Vyi = 0), the equation simplifies to:

Vyf = g * t
Vyf ≈ 9.8 m/s^2 * 8.05 s
Vyf ≈ 79.19 m/s

The angle of the velocity (θ') with respect to the horizontal can be found using:

θ' = arctan(Vyf / Vx)

θ' = arctan(79.19 m/s / 63.957 m/s)
θ' ≈ arctan(1.238)
θ' ≈ 50.486 degrees

The angle of the velocity just before impact is approximately 50.486 degrees below the horizontal.

Answer 2

Therefore, the answers are:

a. The balloon will travel approximately 515.278 meters horizontally before striking the ground.

b. The angle of the velocity just before impact is approximately 50.486 degrees below the horizontal.