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.
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.