Question

A few seconds after a plane took off it was hit by a missile launched from a distance on the ground. The path of the plane is described by y=(0.0005 x?)m. The missile traveled along a parabolic path described by y2 = (160 x), where the coordinates are measured in meters. If the components of the missile velocity in the horizontal and vertical directions are constant at vx =120 m/s and v; = 100 m/s, respectively, and the plane was rising with a constant upward velocity vy = 12 m/s, determine the magnitudes of the velocity and acceleration of the plane the moment it was hit by the missile. у y=(0.0005 x) y2 = (160 x) X

Answer 1

given, path of plane
y = 0.0005x^2

path of missile
y^2 = 160x
given, components of missile velocity in horizontal and vertical directinos are constant at
vx = 120 m/s and vy = 100 m/s

and the plane was rising upward with vy = 12 m/s

now, for the plane
at time of collision, vy = 12 m/s = dy/dt
0.001x*vx = 12 [ where vx is horizontal velocity of the plane at the instant of collision]
x*vx = 12000

for the missile
vx is constant , vx = 120 m/s
vy = constant = 12m/s

so,
2y*dy/dt = 160*dx/dt
2y*vy = 160*vx
2y*100 = 160*120
y = 96 m
  
so, from the equation of motion of the plane
96/0.0005 = x^2
x = 438.1780460

so for the plane
vy = 12 m/s, vx = 12000/x = 12000/438.17804600 = 27.386127875 m/s
  
from the path of plane
vy = 0.001x*vx
ay = 0.001(x*ax + vx^2)

so, ay = 0.001(438.178*ax + 27.386127875^2)
now plane was rising up with constant velocity
so, ay = 0
hence
ax = -27.386127875^2/438.178
ax = -1.7116331719073 m/s/s

hence at the tiem of collision
the plane velocity is v = sqrt(vx^2 + vy^2) = 29.89971565 m/s
direction is arctan(vy/vx) = arctan(12/27.38) = 23.66 deg counterclocwsie above x axis

acceleration is a = -1.711611 m/s/s (-ve direction indicating the acceleratino is in -ve x direction)