An airplane with a speed of 80.9 m/s is climbing upward at an angle of 64.5 ° with respect to the horizontal. When the plane's altitude is 795 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.
a. h = 0.5g*t^2 = 795 m.
4.9*t^2 = 795
t^2 = 162.24
Tf = 12.74 s. = Fall time.
d = 80.9*cos64.5 * 12.74 = 444 m.
To solve this problem, we can break it down into two parts. First, we will calculate the time it takes for the package to hit the ground. Then, using this time, we can determine the horizontal distance traveled by the package. Finally, we will find the angle of the velocity vector just before impact.
(a) Calculating the time taken by the package to hit the ground:
We can calculate the time using the vertical component of the motion. Since the plane is flying at an angle of 64.5° with respect to the horizontal, we can find the vertical component of the airplane's velocity by multiplying its speed (80.9 m/s) by the sine of the angle (64.5°):
Vertical component of velocity = 80.9 m/s * sin(64.5°) = 72.246 m/s
Using this vertical velocity and the initial altitude of 795 m, we can apply the kinematic equation for vertical motion:
Final altitude = Initial altitude + (Vertical component of velocity * time) - (0.5 * acceleration * time^2)
Assuming the acceleration due to gravity is -9.8 m/s^2 (taking downward as negative), we can re-arrange the equation to solve for time:
0 = 795 m + (72.246 m/s * time) - (0.5 * (-9.8 m/s^2) * time^2)
Simplifying this equation, we get a quadratic equation in terms of time:
4.9t^2 - 72.246t - 795 = 0
We can solve this quadratic equation using the quadratic formula. Plugging in the values, we get:
t = (-(-72.246) ± sqrt((-72.246)^2 - 4 * 4.9 * (-795))) / (2 * 4.9)
Simplifying further, we get two possible values for time: t1 ≈ 12.38 s and t2 ≈ -7.89 s. Since time cannot be negative in this context, we discard the negative value.
Therefore, the package takes approximately 12.38 seconds to hit the ground.
(b) Determining the angle of the velocity vector just before impact:
To find the angle, we can use the horizontal and vertical components of the package's velocity just before impact.
The horizontal component remains constant throughout the motion, so we can use the horizontal component of the airplane's velocity (80.9 m/s * cos(64.5°)) as the horizontal component of the package's velocity:
Horizontal component of velocity = 80.9 m/s * cos(64.5°) ≈ 37.038 m/s
The vertical component is determined by the time it takes for the package to hit the ground. We already know the time from part (a), which is approximately 12.38 seconds. So, we can calculate the vertical component using the equation:
Vertical component of velocity = Initial vertical component of velocity + (acceleration * time)
Since the vertical component of velocity at release is 72.246 m/s (calculated in part (a)), and the acceleration due to gravity is -9.8 m/s^2, we get:
Vertical component of velocity ≈ 72.246 m/s + (-9.8 m/s^2) * 12.38 s
The vertical component of velocity is approximately -100.063 m/s.
Using the horizontal and vertical components of velocity, we can determine the angle using the arctangent function:
Angle = arctan(vertical component of velocity / horizontal component of velocity)
Plugging in the values, we get:
Angle ≈ arctan(-100.063 m/s / 37.038 m/s)
Calculating this using a calculator, we find that the angle is approximately -70.651°.
Therefore, the angle of the velocity vector of the package just before impact, relative to the ground, is approximately -70.651°.