An airplane with a speed of 94.9 m/s is climbing upward at an angle of 63.2 ° with respect to the horizontal. When the plane's altitude is 743 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.
To solve this problem, we can use the principles of projectile motion. Let's break it down into two parts:
(a) Calculating the distance along the ground:
In projectile motion, we can separate the motion into horizontal and vertical components. The horizontal component remains constant, while the vertical component is affected by gravity.
First, let's find the time it takes for the package to reach the ground. We can use the formula for vertical motion:
y = y0 + v0y * t + (1/2) * a * t^2
Here,
y = final vertical position of the package = 0 (as it reaches the ground)
y0 = initial vertical position of the package = 743 m
v0y = initial vertical component of velocity = v * sin(θ)
a = acceleration due to gravity = -9.8 m/s^2
t = time
Plugging in the values:
0 = 743 + (94.9 * sin(63.2)) * t - (4.9 * t^2)
Now, let's solve this quadratic equation to find the time it takes for the package to reach the ground.
We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Here,
a = -4.9
b = 94.9 * sin(63.2)
c = 743
Solving this equation, we get two values for t: one positive and one negative. Since negative time doesn't make sense in this context, we discard it.
Now, we have the time it takes for the package to reach the ground. Let's calculate the horizontal distance traveled by the package.
We can use the formula:
x = x0 + v0x * t
Here,
x = distance along the ground
x0 = initial horizontal position of the package = 0 (as it is directly beneath the release point)
v0x = initial horizontal component of velocity = v * cos(θ)
t = time we calculated earlier
Plugging in the values:
x = 0 + (94.9 * cos(63.2)) * t
Calculate the value of x using the previously calculated time, and you'll get the distance along the ground.
(b) Calculating the angle of the velocity vector just before impact:
The angle of the velocity vector just before impact can be found using trigonometry. Since the vertical component of the velocity is affected by gravity, we need to find the horizontal and vertical components of the velocity just before impact.
The horizontal component of velocity remains constant throughout the motion and is given by:
v0x = v * cos(θ)
The vertical component just before impact can be found using the final time we calculated:
vfy = v0y - a * t
Now, we have the horizontal and vertical components of the velocity just before impact. We can use them to calculate the angle using the inverse tangent function:
θ = arctan(vfy / v0x)
Plug in the values and calculate the angle using the formula above.
By following these steps, you can calculate both the distance along the ground and the angle of the velocity vector just before impact.