A plane, diving with constant speed at an angle of 46.0° with the vertical, rele

Question

A plane, diving with constant speed at an angle of 46.0° with the vertical, releases a projectile at an altitude of 830 m. The projectile hits the ground 4.00 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?
m/s

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

(c) What were the horizontal and vertical components of its velocity just before striking the ground?
m/s (horizontal)
m/s (vertical)

Explanation / Answer

a) 830 = vyt + 0.5gt2 = 4vy + 4.9*16

vy = 188 m/s

v = 188/cos46 = 270.6 m/s

b) vx = 270.6sin46 = 194.68 m/s

t = 4s

x = vt = 194.68*4 = 778.7 m

c) vx = 194.68 m/s

vy = v0y + 1t = 188 + 4*9.8 = 227.2 m/s

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