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 75.2 m/s and a 47° above the horizontal when the balloon is released. At the point of release, the plane is at an altitude of 520 m.
(a) How far horizontally, measured from a point directly below the plane's initial position, will the balloon travel before striking the ground?
m
(b) At the point just before balloon strikes the ground, what angle does its velocity make with the horizontal? Give your answer as an angle measured below the horizontal.
(a) How far horizontally, measured from a point directly below the plane's initial position, will the balloon travel before striking the ground?
Well, it seems our stunt pilot is about to bring a classic party game to the skies! Let's help them out.
To calculate the horizontal distance traveled by the balloon, we need to find the time it takes to hit the ground and then multiply it by the horizontal velocity of the plane.
First, let's find the vertical component of the balloon's velocity. We can use basic trigonometry for this. The vertical velocity (Vy) can be calculated as:
Vy = velocity * sin(angle)
where the velocity is 75.2 m/s and the angle is 47°.
Vy = 75.2 m/s * sin(47°)
Vy ≈ 75.2 m/s * 0.7314
Vy ≈ 54.958 rounding to 3 decimal places.
Now, we can use the vertical motion equation to find the time it takes for the balloon to hit the ground. The equation for vertical motion is:
h = Vi * t + 0.5 * a * t^2
where h is the height (520 m), Vi is the initial vertical velocity (Vy), t is the time, and a is the acceleration (which we assume to be -9.8 m/s^2 due to gravity).
Using some rearranging of this equation, we find:
0.5 * a * t^2 + Vi * t - h = 0
Plugging in the values:
0.5 * (-9.8 m/s^2) * t^2 + 54.958 m/s * t - 520 m = 0
Now, let's use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 0.5 * (-9.8 m/s^2), b = 54.958 m/s, and c = -520 m. Let's plug in these values and solve:
t = (-54.958 ± √(54.958^2 - 4 * 0.5 * (-9.8) * (-520))) / (2 * 0.5 * (-9.8))
Calculating this gives us two solutions for t: t ≈ 9.849 and t ≈ -10.734.
Since time cannot be negative in this context, we'll use the positive value:
t = 9.849 s
Finally, we can calculate the horizontal distance traveled (d) using:
d = velocity * cos(angle) * time
Plugging in the values:
d = 75.2 m/s * cos(47°) * 9.849 s
d ≈ 75.2 m/s * 0.6820 * 9.849 s
d ≈ 502.526 rounding to 3 decimal places.
So, the balloon will travel approximately 502.526 meters horizontally.
(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 below the horizontal.
Well, this balloon is about to go out with a bang! To find the angle, we can use trigonometry again.
The vertical component of the velocity just before the balloon strikes the ground will be equal to the initial vertical velocity (Vy), but in the opposite direction due to the effect of gravity.
So, Vy = -54.958 m/s.
Now we can find the angle, which can be determined using the tangent function:
tan(angle) = Vy / velocity
Substituting in the values:
tan(angle) = -54.958 m/s / 75.2 m/s
angle = arctan(-54.958 / 75.2)
Calculating this gives us an angle of approximately -37.942°, measured below the horizontal.
So, the angle that the balloon's velocity makes with the horizontal just before it strikes the ground is approximately -37.942°, and the balloon will create quite a splash roughly 502.526 meters away!
To find the horizontal distance the balloon will travel, we can use the equation for horizontal distance along with the initial velocity in the horizontal direction and time of flight. Let's find out the initial horizontal velocity first.
The initial horizontal velocity of the balloon can be determined using the given speed of the airplane and the launch angle. We can calculate it using the trigonometric function cosine:
Horizontal velocity (Vx) = Velocity × cos(θ)
= 75.2 m/s × cos(47°)
Now, let's calculate the initial horizontal velocity (Vx):
Vx = 75.2 m/s × cos(47°)
= 75.2 m/s × 0.682
Vx ≈ 51.3 m/s
The balloon will continue to move horizontally at this velocity until it hits the ground.
(a) To find the horizontal distance traveled by the balloon, we can use the equation:
Distance = Velocity × Time
Time can be calculated using the vertical motion of the balloon. We know that the initial vertical velocity is 0 m/s (since it is just being released) and the acceleration due to gravity is -9.8 m/s². The vertical displacement is -520 m (negative since the balloon is falling downwards). We can use the kinematic equation:
Displacement = (Initial Velocity × Time) + (0.5 × Acceleration × Time²)
Plugging in the values, we can solve for time:
-520 m = (0 × Time) + (0.5 × (-9.8 m/s²) × Time²)
-520 m = -4.9 m/s² × Time²
Time² = -520 m / -4.9 m/s²
Time² = 106.12 s²
Time = √(106.12 s²)
Time ≈ 10.3 s
Now, we can calculate the horizontal distance traveled by the balloon using the horizontal velocity (Vx) and time (t):
Distance = Velocity × Time
Distance = 51.3 m/s × 10.3 s
Distance ≈ 528.39 m
Therefore, the horizontal distance the balloon will travel before striking the ground is approximately 528.39 meters.
(b) At the point just before the balloon strikes the ground, its vertical velocity will be -9.8 m/s (acceleration due to gravity) and its horizontal velocity will be 51.3 m/s.
To find the angle the velocity makes with the horizontal, we can use the trigonometric function tangent:
Tan(θ) = Vertical Velocity / Horizontal Velocity
θ = Tan^(-1) (Vertical Velocity / Horizontal Velocity)
θ = Tan^(-1) (-9.8 m/s / 51.3 m/s)
θ ≈ -10.9°
Therefore, at the point just before the balloon strikes the ground, the angle its velocity makes with the horizontal is approximately -10.9 degrees (measured below the horizontal).
To solve this problem, we can use kinematic equations that relate the initial velocity, acceleration, time, and displacement. Let's break down the problem step by step.
(a) To find how far horizontally the balloon will travel before striking the ground, we need to find the time it takes for the balloon to hit the ground. We can use the vertical motion of the balloon for this calculation.
Using the kinematic equation for vertical motion:
h = 1/2 * g * t^2
Where:
h is the height (in this case, 520m),
g is the acceleration due to gravity (approximately 9.8 m/s²), and
t is the time.
Rearranging the equation to solve for t:
t = sqrt(2h / g)
Plugging in the given values:
t = sqrt(2 * 520m / 9.8 m/s^2)
t ≈ 10.0901 seconds
Now that we have the time, we can find the horizontal distance. Using the horizontal motion of the balloon, we can determine the x-component of the initial velocity.
vx = v * cosθ
Where:
v is the initial velocity (75.2 m/s), and
θ is the launch angle (47° above the horizontal).
Plugging in the given values:
vx = 75.2 m/s * cos(47°)
vx ≈ 50.7968 m/s
Now, we can multiply the horizontal velocity by the time to find the horizontal distance:
d = vx * t
d ≈ 50.7968 m/s * 10.0901 s
d ≈ 512.6095 meters
Therefore, the balloon will travel approximately 512.61 meters horizontally before striking the ground.
(b) To find the angle of the velocity just before the balloon strikes the ground, we need to find the vertical and horizontal components of the velocity at that moment.
Using the horizontal and vertical components of the initial velocity:
vx = v * cosθ
vy = v * sinθ
Using the kinematic equation for horizontal motion:
dx = vx * t
Using the kinematic equation for vertical motion:
dy = vy - 1/2 * g * t^2
We know that at the point just before the balloon strikes the ground, dy = 0 (since the vertical displacement becomes zero). Therefore:
0 = vy - 1/2 * g * t^2
Rearranging the equation to solve for vy:
vy = 1/2 * g * t^2
Plugging in the given values:
vy = 1/2 * 9.8 m/s² * (10.0901 s)^2
vy ≈ 494.3427 m/s
To find the angle measured below the horizontal, we can use the inverse tangent:
θ' = arctan(vy / vx)
Plugging in the given values:
θ' = arctan(494.3427 m/s / 50.7968 m/s)
θ' ≈ 83.5807°
Therefore, the angle of the velocity just before the balloon strikes the ground is approximately 83.58° below the horizontal.
Vo = 75.2m/s @ 47o
Xo = 75.2*cos47 = 51.3 m/s.
Yo = 75.2*sin47 = 55.0 m/s.
h = 520 m
a. Yo*t + 0.5g*t^2 = 520 m.
-55*t + 4.9*t^2 = 520
4.9t^2 - 55t -520 = 0
Use Quadratic formula.
Tf = 17.34 s. = Fall time.
Dx = Xo*Tf = 51.3m/s * 17.34s = 890 m
b. Y = Yo + g*t = -55 + 9.8*17.34 = 115 m/s.
V = Xo + Yi = 51.3 + 115i
Tan A = Y/Xo = 115/51.3 = 2.24039
A = 66o or -66o = 66o Below the +x axis.