Question
With the aid of diagrams and mathematical calculations, solve:
An object is released from an airplane which is driving at an angle of 30 degrees Celsius from the horizontal with a speed of 50m/s. If the plane is at a height of 4000m from the ground when the object is released, find;
(A). The velocity of the object when it hits the ground.
(B). The time taken for the object to hit the ground.
An object is released from an airplane which is driving at an angle of 30 degrees Celsius from the horizontal with a speed of 50m/s. If the plane is at a height of 4000m from the ground when the object is released, find;
(A). The velocity of the object when it hits the ground.
(B). The time taken for the object to hit the ground.
Answers
GPT 3.5
Let's first draw a diagram to visualize the situation:
*
|
| 30°
|
airplane ---- *
|
| 4000 m
*
|
|
GND
In this diagram, the airplane is represented by the line segment "* - *", and the ground is represented by the line segment "* - GND" (the horizontal line).
We can split the initial velocity of the object into horizontal and vertical components. The horizontal component is given by v₀x = v₀ * cosθ, where v₀ = 50 m/s and θ = 30°. The vertical component is given by v₀y = v₀ * sinθ.
(A). The velocity of the object when it hits the ground:
The vertical motion of the object is influenced by gravity, so we can use the kinematic equation:
y = y₀ + v₀y * t - 0.5 * g * t², where y₀ = 4000 m, v₀y = v₀ * sinθ, and g = 9.8 m/s².
At the moment the object hits the ground, y = 0. Therefore, we can solve the equation for t:
0 = 4000 + (50 * sin30°) * t - 0.5 * 9.8 * t²
Simplifying, we get:
0.5 * 9.8 * t² - 25t - 4000 = 0
Using the quadratic formula, t = (-b ± √(b² - 4ac)) / (2a), where a = 0.5 * 9.8, b = -25, and c = -4000, we can solve for t.
t = (-(-25) ± √((-25)² - 4 * 0.5 * 9.8 * -4000)) / (2 * 0.5 * 9.8)
t = (25 ± √(625 + 78400)) / 9.8
Taking the positive root (we don't consider negative time), we find:
t ≈ 82.3 s
Now, for the horizontal motion, we can use the equation:
x = x₀ + v₀x * t, where x₀ = 0 (starting point) and v₀x = v₀ * cosθ.
Substituting the known values, we get:
x = 0 + (50 * cos30°) * 82.3
x ≈ 42.5 * 82.3
x ≈ 3499.75 m
Thus, the object hits the ground at approximately x = 3499.75 m horizontally.
(B). The time taken for the object to hit the ground:
From the above calculations, we found that it takes approximately t = 82.3 s for the object to hit the ground.
*
|
| 30°
|
airplane ---- *
|
| 4000 m
*
|
|
GND
In this diagram, the airplane is represented by the line segment "* - *", and the ground is represented by the line segment "* - GND" (the horizontal line).
We can split the initial velocity of the object into horizontal and vertical components. The horizontal component is given by v₀x = v₀ * cosθ, where v₀ = 50 m/s and θ = 30°. The vertical component is given by v₀y = v₀ * sinθ.
(A). The velocity of the object when it hits the ground:
The vertical motion of the object is influenced by gravity, so we can use the kinematic equation:
y = y₀ + v₀y * t - 0.5 * g * t², where y₀ = 4000 m, v₀y = v₀ * sinθ, and g = 9.8 m/s².
At the moment the object hits the ground, y = 0. Therefore, we can solve the equation for t:
0 = 4000 + (50 * sin30°) * t - 0.5 * 9.8 * t²
Simplifying, we get:
0.5 * 9.8 * t² - 25t - 4000 = 0
Using the quadratic formula, t = (-b ± √(b² - 4ac)) / (2a), where a = 0.5 * 9.8, b = -25, and c = -4000, we can solve for t.
t = (-(-25) ± √((-25)² - 4 * 0.5 * 9.8 * -4000)) / (2 * 0.5 * 9.8)
t = (25 ± √(625 + 78400)) / 9.8
Taking the positive root (we don't consider negative time), we find:
t ≈ 82.3 s
Now, for the horizontal motion, we can use the equation:
x = x₀ + v₀x * t, where x₀ = 0 (starting point) and v₀x = v₀ * cosθ.
Substituting the known values, we get:
x = 0 + (50 * cos30°) * 82.3
x ≈ 42.5 * 82.3
x ≈ 3499.75 m
Thus, the object hits the ground at approximately x = 3499.75 m horizontally.
(B). The time taken for the object to hit the ground:
From the above calculations, we found that it takes approximately t = 82.3 s for the object to hit the ground.