Please only answer if you are sure ...second time asking....A plane, diving with

Question

Please only answer if you are sure ...second time asking....A plane, diving with constant speed at an angle of 52.0° with the vertical, releases a projectile at an altitude of 570 m. The projectile hits the ground 8.00 s after release. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction. Neglect air resistance.) (a) What is the speed of the aircraft?
52.1 m/s

(b) How far did the projectile travel horizontally during its flight?
328.4 m

(c) What were the horizontal and vertical components of its velocity just before striking the ground? (Indicate the direction with the sign of your answer.)

Explanation / Answer

a) let the speed of plane v, then the downward component of its velocity is vcos(52)=d.

Then, at the beginning, the projectile has downward velocity d, and at the end, it has downward velocity d+8*9.81 = d+78.48. That means its average downward velocity is (d + d +78.48)/ 2 =d+39.24. It falls a total of 570m,

so 570 = (d+39.24)*(8).
d+39.24 = 71.25 => d = 32.01 m/s.
Then, v = d / cos(52) = 52 m/s.

b) the horizontal velocity is 52*sin(42.3) = 41m/s. This velocity is horizontal, so it's not accelerated by gravity. So, the horizontal distance is 41m/s*8s = 328m

c) We just said that the magnitute of the horizontal component didn't change. It was 41 m/s.

d) Then we recall that the final downward velocity was d+39.24 = 41+39.24 = 80.24m/s.

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