A plane, diving with constant speed at an angle of 53.0 degree with the vertical

Question

A plane, diving with constant speed at an angle of 53.0 degree with the vertical, releases a projectile at an altitude of 500 m. The projectile hits the ground 6.00 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? m/s How far did the projectile travel horizontally during its flight? m What were the horizontal and vertical components of its velocity just before striking the ground? m/s (horizontal) m/s (vertical)

Explanation / Answer

here,

angle, A = 53 degrees
height of plane, h = 500 m
time to hit ground, t = 6 s

Part A:
In projectile motion,
time of flight, t = 2v*SinA/g

Rewriting for velocity, v = g*t/(2*sinA)
v = 9.81*6/(2*Sin53)
v = 36.85 m/s

Part B:
Horizontal Distance is given by expression,
X = v^2*Sin2A/g
X = 36.85^2*Sin(2*53)/(9.81)
X = 133.060 m

Part C:
Horizontal Velocity, Vx = v*Cos53 = 53*Cos53 = 31.896 m/s

Vertical Velocity, Vy = v*Sin53 = 53*Sin53 = 42.327 m/s

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