A plane, diving with constant speed at an angle of 55.0° with the vertical, rele

Question

A plane, diving with constant speed at an angle of 55.0° with the vertical, releases a projectile at an altitude of 590 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.)

What is the speed of the aircraft?

How far did the projectile travel horizontally during its flight

What were the horizontal and vertical components of its velocity just before striking the ground?

Explanation / Answer

= 55.0°, h = 590 m, t = 8.00 s, find the speed of the aircraft v.

initial vertical velocity = -vcos

vertical displacement = -h = (-vcos)t - gt2/2

v = (h/t - gt/2)/cos = 60.2 m/s

How far did the projectile travel horizontally during its flight? find d

d = vsin*t = 395 m

What were the horizontal and vertical components of its velocity just before striking the ground?

vx' = vsin = 49.3 m/s

vy' = vcos - gt = -43.9 m/s

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