An airplane with a speed of 74.5 m/s is climbing upward at an angle of 64.8 degrees with respect to the horizontal. When the plane's altitude is 687 m, the pilot releases the 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.

Vertical speed = 74.5 sin 64.8 = 67.40961541

Extra height reached = 4544.05625 / (2 x 9.8) = 231.8396046

Total height reached = 231.8396046 + 687 = 918.8396046 m

Time to reach max height = 74.5/9.8 = 7.602040816 s

Time to fall 919 m = square root (2 x 918.8396046 / 9.8) = 13.69373166 s

Total time in air = 21.29577248 seconds

Horizontal speed = 74.5 cos 64.8 = 31.72055722 m/s

Horizontal distance = 31.72055722 x 21.29577248 = 675.5137695 m

Vertical speeds at ground = 9.8 x 13.69373166 = 134.1985703 m/s

Angle to the ground velocity = tan^-1 (134.1985703 / 31.72055722) = 76.70106771 degrees

For part A my answer is 134.1985703 m/s
For part B my answer is 76.70106771 degrees
However, the answers are incorrect. What am I doing wrong? Can you please help. Thanks.

Let's go through the problem step by step to identify any mistakes.

1. First, let's calculate the vertical component of the airplane's velocity. The vertical speed is given by V_vertical = V * sin(θ), where V is the airplane's speed and θ is the angle of climb. Plugging in the values, we get V_vertical = 74.5 * sin(64.8°) = 67.38 m/s (rounded to two decimal places). It seems you rounded to 67.41 m/s, which is slightly inaccurate.

2. Next, let's calculate the extra height reached by the airplane. To do this, we can use the formula h = V_vertical * t - (1/2) * g * t^2, where h is the extra height, t is the time taken to reach that height, and g is the acceleration due to gravity (approximately 9.8 m/s^2). The time taken to reach the maximum height can be found using t = V_vertical / g. Plugging in the values, we get t = 67.38 / 9.8 = 6.88 s (rounded to two decimal places).

Substituting this time into the formula, we get h = 67.38 * 6.88 - (1/2) * 9.8 * 6.88^2 ≈ 267.22 m (rounded to two decimal places). It seems you calculated 4544.06 m, which seems to be an incorrect substitution.

3. Now let's calculate the total height reached by the airplane. The total height is the sum of the extra height and the initial altitude. So, the total height reached is 918.8396 m (as you correctly calculated).

4. Next, let's calculate the time taken for the package to fall from the total height. We can use the formula h = (1/2) * g * t^2 to solve for t. Rearranging the formula, we get t = sqrt(2h / g), where sqrt represents the square root. Plugging in the total height, we have t = sqrt(2 * 918.8396 / 9.8) ≈ 13.72 s (rounded to two decimal places). It seems you rounded to 13.69 s, which is slightly inaccurate.

5. The total time in the air is the sum of the time taken to reach the maximum height and the time taken to fall back down. So, the total time in the air is 6.88 + 13.72 ≈ 20.6 s (rounded to one decimal place). It seems you calculated 21.30 s, which is slightly inaccurate.

6. Now, let's calculate the horizontal distance traveled by the airplane during the total time in the air. We can use the formula d = V_horizontal * t, where V_horizontal is the horizontal component of the airplane's velocity. Plugging in the values, we get d = 74.5 * cos(64.8°) * 20.6 ≈ 1487.10 m (rounded to two decimal places). It seems you calculated 675.51 m, which is incorrect.

7. Finally, let's calculate the angle of the velocity vector of the package just before impact relative to the ground. We can use the formula tan^(-1)(V_vertical_final / V_horizontal_final), where V_vertical_final is the vertical velocity just before impact (which is solely due to gravity) and V_horizontal_final is the horizontal velocity just before impact. Since V_vertical_final = -g * t (where g is the acceleration due to gravity and t is the total time in the air), and V_horizontal_final = V_horizontal (which remains constant), we can use these values to calculate the angle. Plugging in the values, we get tan^(-1)(-9.8 * 20.6 / 74.5) ≈ -9.48° (rounded to two decimal places).

Therefore, the corrected answers are:

(a) The distance along the ground measured from a point directly beneath the point of release to where the package hits the earth is approximately 1487.10 meters.

(b) The angle of the velocity vector of the package just before impact, relative to the ground, is approximately -9.48°. Please note that the negative sign indicates that the velocity vector is pointing downwards.

I hope this helps! Let me know if you have any further questions.