A plane, diving with constant speed at an angle of 42.3° with the vertical, rele

Question

A plane, diving with constant speed at an angle of 42.3° with the vertical, releases a projectile at an altitude of 601 m. The projectile hits the ground 5.24 s after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (State your answers to (c) and (d) as positive numbers.)

Im not sure how to work this one at all.

Explanation / Answer

a) If the plane has speed v, then the downward component of its velocity is vcos(42.3)=d. (I introduced the variable d for simplicity.)

Then, at the beginning, the projectile has downward velocity d, and at the end, it has downward velocity d+5.24*9.81 = d+51.4044. That means its average downward velocity is (d + d + 51.4044)/ 2 =d+25.7022. It falls a total of 601m, so 601 = (d+25.7022)(5.24).
d+25.7022 = 114.69.... d= 88.992 m/s.
Then, v = d / cos(42.3) = 120.32 m/s.

b) the horizontal velocity is 120.32*sin(42.3) = 80.98m/s. This velocity is horizontal, so it's not accelerated by gravity. So, the horizontal distance is 80.98m/s* 5.24s=424.32m

c) We just said that the magnitute of the horizontal component didn't change. It was 80.98m/s.

d) Then we recall that the final downward velocity was d+51.4044=88.992+51.4044=140.40m/s.

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